Skip to main content

Full text of "Calculus"

See other formats


THE LIBRARY 

OF 

THE UNIVERSITY 

OF CALIFORNIA 

LOS ANGELES 



Tk RALPfl D. REED LIBRART 

DEPARTMENT OP GHKLOGT 

ijMivKKarrr ^ California 

UM ANGCLEB. CAUF, 



Digitized by tine Internet Arciiive 

in 2007 witii funding from 

IVIicrosoft Corporation 



littp://,www.arcliive.org/details/calculusOOmarciala 



CALCULUS 



MODERN MATHEMATICAL TEXTS 

EDITED BY 

Charles S. Slighter 



ELEMENTARY MATHEMATICAL ANALYSIS 
By Charles S. Slighter 
490 pages, 5 x IM, Illustrated. 

MATHEMATICS FOR AGRICULTURAL 
STUDENTS 
By Henry C. Wolff 
311 pages, 5 x 7J.i, Illustrated . 

CALCULUS 
By Herman W. March and Henry C. "Wolff 
360 pages, 5 x TVi, Illustrated . 



MODERN MATHEMATICAL TEXTS 
Edited by Charles S. Slighter 



CALCULUS 



BY 
HERMAN W. MARCH, Ph. D. 



ASSOCIATE PKOFE880R OP MATHEMATICS 
UNIVEKSITY OF WISCONSIN 



AND 



HENRY C. WOLFF, Ph. D. 

PBOFESSOR OF MATHEMATICS 
DREXEIi INSTITUTE 



First Edition 
Sixth Impression 



McGRAW-HILL BOOK COMPANY, Inc. 
NEW YORK: 370 SEVENTH AVENUE 

LONDON: 6 & 8 BOUVERIE ST., E. C. 4 
1917 



Copyright, 1917, by the 
McGraw-Hill Book Company, Inc. 



THE MAPLE PRESS YORK PA 



Geolo^^ 
Library 



PREFACE 



One of the purposes of the elementary working courses in mathe- 
matics of the freshman and sophomore years is to exhibit the bond 
that unites the experimental sciences. "The bond of union 
among the physical sciences is the mathematical spirit and the 
mathematical method which pervade them." For this reason, the 
applications of mathematics, not to artificial problems, but to the 
more elementary of the classical problems of natural science, find 
a place in every working course in mathematics. This presents 
probably the most difficult task of the text-book writer, — namely, 
to make clear to the student that mathematics has to do with the 
laws of actual phenomena, without at the same time undertaking 
to teach technology, or attempting to build upon ideas which the 
student does not possess. It is easy enough to give examples of 
the application of the processes of mathematics to scientific prob- 
lems; it is more difficult to exhibit by these problems, how, in 
mathematics, the very language and methods of thought fit 
naturally into the expression and derivation of scientific laws and of 
natural concepts. 

It is in this spirit that the authors have endeavored to develop 
the fundamental processes of the calculus which play so important 
a part in the physical sciences; namely, to place the emphasis upon 
the mode of thought in the hope that, even though the student may 
forget the details of the subject, he will continue to apply these 
fundamental modes of thinking in his later scientific or technical 
career. It is with this purpose in mind that problems in geometry, 
physics, and mechanics have been freely used. The problems 
chosen will be readily comprehended by students ordinarily taking 
the first course in the calculus. 

A second purpose in an elementary working course in mathe- 
matics is to secure facility in using the rules of operation which 
must be applied in calculations. Of necessity large numbers of 
drill problems have been inserted to furnish practice in using the 



104^284 



vi PREFACE 

rules. It is hoped that the solution of these problems will be re- 
garded by teacher and student as a necessary part but not the 
vital part of the course. 

While the needs of technical students have been particularly in 
the minds of the authors, it is believed that the book is equally 
adapted to the needs of any other student pursuing a first course 
in calculus. The authors do not believe that the purposes of 
courses in elementary mathematics for technical students and for 
students of pure science differ materially. Either of these classes 
of students gains in mathematical power from the type of study 
that is often assumed to be fitted for the other class. 

In agreement with many others, the book is not divided into two 
parts. Differential Calculus and Integral Calculus. Integration 
with the determination of the constant of integration, and the 
definite integral as the limit of a sum, are given immediately fol- 
lowing the differentiation of algebraic functions and before the 
differentiation of the transcendental functions. With this arrange- 
ment many of the most important applications of the calculus 
occur early in the course and constantly recur. Further, with this 
arrangement, the student is enabled to pursue more advantageously 
courses in physics and mechanics simultaneously with the calculus. 

The attempt has been made to give infinitesimals their proper 
importance. In this connection Duhamel's Theorem is used as a 
valuable working principle, though the refinements of statement 
upon which a rigorous proof can be based have not been given. 

The subjects of center of gravity and moments of inertia have 
been treated somewhat more fully than is usual. They are par- 
ticularly valuable in emphasizing the concept of the definite 
integral as the limit of a sum and as a mode of calculating the 
mean value of a function. Sufficient solid analytic geometry is 
given to enable students without previous knowledge of this sub- 
ject to work the problems involving solids. In the last chapter 
simple types of differential equations are taken up. 

The book is designed for a course of four hours a week through- 
out the college year. But it is easy to adapt it to a three-hour 
course by suitable omissions. 

The authors are indebted to numerous current text-books for 
many of the exercises. To prevent distracting the student's at- 



PREFACE vii 

tention from the principles involved, exercises requiring compli- 
cated reductions have been avoided as far as possible. 

The book in a preliminary form has been used for two years 
with students in the College of Engineering of the University of 
Wisconsin. Many improvements have been suggested by our 
colleagues, Professor H. T. Burgess, Messrs. E. Taylor, T. C. Fry, 
J. A. Nyberg, and R. Keffer. Particular acknowledgment is due 
to the editor of this series. Professor C. S. Slichter, for suggestions 
as to the plan of the book and for suggestive criticism of the manu- 
script at all stages of its preparation. 

The authors will feel repaid if a little has been accomplished 
toward presenting the calculus in such a way that it will appeal to 
the average student rather as a means of studying scientific prob- 
lems than as a collection of proofs and formulas. 

Unfversity op Wisconsin, Herman W. March, 

November 6, 1916. Henry C, Wolff. 



CONTENTS 

Page 

Preface v 

Introduction 1 

1. Constant. Variable. Function . . - 1 

2. The Power Function 1 

3. The Law of the Power Function 3 

4. Polynomials. Algebraic Function 4 

5. Transcendental Functions " . 6 

6. Translation 6 

7. Elongation or Contraction, or Orthographic Projection, of a 
Curve . 7 

8. Shear 8 

9. The Function a' 9 

10. The Function sin x 9 

11. The Functions p = a cos 0, p = b sin 0, and p = a cos d + 

6 sin e 9 

12. Fundamental Transformations of Functions 10 

CHAPTER I 

Derivative 

13. Increments 13 

14. The Function y = x^ 16 

15. Slope of the Tangent 17 

16. Maxima and Minima 20 

17. Derivative 21 

18. Velocity of a Falling Body 22 

19. Illustrations 23 

20. Illustrations 23 

CHAPTER II 

Limits 

21. Definition 27 

22. Notation 29 



X CONTENTS 

Section Page 

23. Infinitesimal 29 

24. Theorems on Limits 30 

25. The Indeterminate Form ^ 30 

26. Continuous and Discontinuous Functions 31 

CHAPTER III 

Thf Power Function 

27. The Power Function 33 

28. The Derivative of x» 33 

29. The Derivative of ax" 35 

30. Rate of Change of ax" 37 

31. The Derivative of the Sum of a Function and a Constant . 39 

32. The Derivative of au" 41 

33. The Derivative of u", n a Positive Fraction 42 

34. The Derivative of a Constant 43 

35. The Derivative of the Sum of Two Functions 43 

36. Differentiation of ImpHcit Functions 44 

37. Anti-derivative. Integration 46 

38. Acceleration 52 

CHAPTER IV 

DiPFEKENTIATION OF AlGBBRATION FUNCTIONS 

39. The Derivative of the Product of a Constant and a Variable. 56 

40. The Derivative of the Product of Two Functions 57 

41. The Derivative of the Quotient of Two Functions .... 58 

42. The Derivative of w", n negative 59 

43. Maximum and Minimum Values of a Function 00 

44. Derivative of a Function of a Function 66 

45. Inverse Functions 67 

46. Parametric Equations 67 

47. Lengths of Tangent, Normal, Subtangent, and Subnormal 68 

CHAPTER V 
Second Derivative. Point of Inflection 

48. Second Derivative. Concavity 70 

49. Points of Inflection 71 



CONTENTS xi 

CHAPTER VI 

Applications 
Section Page 

50. Area Under Curve: Rectangular Coordinates 75 

51. Work Done by a Variable Force 77 

52. Parabolic Cable 82 

53. Acceleration 83 

54. The Path of a Projectile 85 

CHAPTER VII 

Infinitesimals. Differentials. Definite Integrals 

55. Infinitesimals 88 

56. j;'!'.'^ • ■ ■ ■ 88 

lim tan a lim tan a 
^'- a = Q~^'(x^Q^vr~a. ^^ 

58. ^^ l^^^ 89 

a: = a 

59. Order of Infinitesimals 90 

60. Theorem 93 

61. Differentials 95 

62. Formulas for the Differentials of Functions 96 

63. Differential of Length of Arc : Rectangular Coordinates . . 100 

64. The Limit of 2;/(a;)Aa; 102 

65. Definite Integral 103 

66. Duhamel's Theorem 105 

•67. Work Done by a Variable Force 107 

68. Volume of a Solid of Revolution 109 

69. Length of Arc: Rectangular Coordinates . Ill 

70. Area of a Surface of Revolution 112 

71. Element of Integration 113 

72. Water Pressure 114 

73. Airthmetic Mean 116 

74. Mean Value of a Function 117 

CHAPTER VIII 

Circular Functions. Inverse Circular Functions 

75. The Derivative of sin m 122 

76. The Derivative of cos u, tan m, cot u, sec m, and esc u . .124 



xii CONTENTS 

Section Page 

77. The Derivative of the Inverse Circular Functions .... 131 

78. Velocity and Acceleration 136 

79. Angular Velocity and Acceleration 139 

80. Simple Harmonic Motion 140 

81. The Simple Pendulum 141 

CHAPTER IX 

Exponential and Logarithmic Functions 

82. The Derivative of the Exponential and Logarithmic Func- 
tions 145 

83. Logarithmic Differentiation 153 

84. Compound Interest Law 155 

85. Relative Rate of Increase 159 

86. Hyperbolic Function 159 

87. Inverse Hyperbolic Functions 160 

88. The Catenary 161 

CHAPTER X 

Maxima and Minima 

89. The Maximum or Minimum of y = ax^ + fix + y . . . . 165 

90. The Function a cos x + b sin x 165 

91. The Function mx ± \^a^ - s" 168 

92. Ma.xima and Minima by Limits of Curve 169 

93. Maxima and Minima Determined by the Derivative . . . 169 

94. Second-derivative Test for Maxima and Minima 172 

95. Study of a Function by Means of its Derivatives 172 

96. Applications of Maxima and Minima 174 

CHAPTER XI 

Polar Coordinates 

97. Direction of Curve in Polar Coordinates 178 

98. Differential of Arc : Polar Coordinates 181 

99. Area: Polar Coordinates 183 



CONTENTS xiii 

CHAPTER XII 

Integration 

Section Page 

100. Formulas 185 

101. Integration of Expressions Containing ax^ + bx -\- c, by 
Completing the Square 190 

102. Integrals Containing Fractional Powers of a: or of a + fca; . 192 

103. Integrals of Powers of Trigonometric Functions 193 

104. Integration of Expressions Containing \/a^ — x^,'\/a'^ + x^, 
'\/x^ — o*, by Trigonometric Substitution 196 

105. Change of Limits of Integration 199 

106. Integration by Parts 201 

.07.Tho.„tegraU/e«si„n...,/.«c«„... .202 

108. J sm3 xdx 204 

109. Wallis' Formulas 204 

,^-, ., , ,. , Ta sin X + 6 cos X , 

110. Integration of I — -. -. — j ax 209 

J c sm x -\- a cos x 

111. Partial Fractions 211 

CHAPTER XIII 

Applications of the Process of Integration. Improper 
Integrals 

112. Summary of Applications 208 

113. Improper Integrals 223 

114. Improper Integral: Infinite Limits 226 

CHAPTER XIV 
Solid Geometry 

115. Coordinate Axes. Coordinate Planes 228 

116. The Distance Between Two Points 230 

117. Direction Cosines of a Line 230 

118. Angle Between Two Lines 231 

119. The Normal Form of the Equation of a Plane 232 

120. The Equation Ax + By + Cz == D 233 

121. Intercept Form of the Equation of a Plane . 234 

122. The Angle Between Two Planes 236 

123. Parallel and Perpendicular Planes 236 



xiv CONTENTS 

Section Page 

124. The Distance of a Point From a Plane 237 

125. Symmetrical Form of the Equation of a Line 238 

126. Surface of Revolution 240 

127. Quadric Surfaces 242 

128. Cylindrical Surfaces 244 

129. Partial Derivatives 246 

130. Partial Derivatives of Higher Order 247 

CHAPTER XV 

SuccEssrvE Integration. Center of Gravity. Moment 
OF Inertia 

131. Introduction 250 

132. Illustration of Double Integration 251 

133. Area by Double Integration : Rectangular Coordinates . . . 253 

134. Geometrical Meaning of the Definite Double Integral . . . 255 

135. Area: Polar Coordinates 258 

136. Volume of a Solid: Triple Integration 260 

137. Center of Mass. Centroid 263 

138. Centroid Independent of the Position of the Coordinate 
Planes 264 

139. Center of Gravity . 264 

140. Centroid of a Continuous Mass 265 

141. Theorems of Pappus 274 

142. Centroid: Polar Coordinates 276 

143. Moment of Inertia . . . ' 277 

144. Transfer of Axes 280 

145. Moment of Inertia of an Area 281 

146. Moment of Inertia: Polar Coordinates 283 

147. Moment of Inertia of a Solid 284 

CHAPTER XVI 

Curvature. Evolutes. Envelopes 

148. Curvature 289 

149. Curvature of a Circle 290 

150. Circle of Curvature. Radius of Curvature. Center of 
Curvature 290 

151. Formula for Curvature and Radius of Curvature: Rectan- 
gular Coordinates 290 



CONTENTS XV 

Section Page 

152. Curvature: Parametric Equations 292 

153. Approximate Formula for the Curvature 293 

154. Center of Curvature. Evolute 294 

155. Envelopes 296 

156. The Evolute as the Envelope of the Normals 300 

157. Involutes 302 

CHAPTER XVII 

Series. Taylor's and Maclaurin's Theorems. Indeter- 
minate Forms 

158. Infinite Series 303 

159. Rolle's Theorem 305 

160. Law of the Mean 306 

161. The Extended Law of the Mean 307 

162. Taylor's Theorem with the Remainder 308 

163. Taylor's and Maclaurin's Series 311 

164. Second Proof for Taylor's and Maclaurin's Series .... 313 

165. Tests for the Convergence of Series 315 

166. Computation of Logarithms 320 

167. Computation of tt 321 

168. Relation Between the Exponential and Circular Functions 321 

169. DeMoivre's Theorem 323 

170. Indeterminate Forms 324 

CHAPTER XVIII 
Total- Derivative. Exact Differential 

171. The Total Derivative 328 

172. Exact Differential 332 

173. Exact Differential Equations 335 

174. Envelopes .336 

CHAPTER XIX 

Differential Equations 

175. Differential Equations 338 

176. General Solution. Particular Solution 338 

177. Exact Differential Equation 339 

178. Differential Equations: Variable Separable 339 



xvi CONTENTS 

Section Page 

179. Homogeneous Differential Equations 340 

180. Linear Differential Equations of the First Order 342 

181. Extended Form of the Linear Differential Equation . . . 344 

182. Applications 345 

183. Linear Differential Equations of Higher Order with Con- 
stant Coefficients and Second Member Zero 347 

184. Au.xiliary Equation with Equal Roots 349 

185. Auxiliary Equation with Complex Roots 350 

186. Damped Harmonic Motion 351 

Index 154 



CALCULUS 



INTRODUCTION 

1. Constant. Variable. Function. 1. A symbol of number 
or quantity, as a, to which a fixed value is assigned throughout 
the same problem or discussion is called a constant. 

2. A symbol of number or quantity, as x, to which a succession 
of values is assigned in the same problem or discussion is called a 
variable. 

Example. The mass or weight of mercury in a thermometer is 
constant. The number that results from measuring this quantity 
(weight) is a constant. 

The volume of the mercury in the thermometer is variable. 
The number that results from measuring this quantity (volume) 
is a variable. 

3. The variable y is said to be a function of the variable x if, 
when X is given, one or more values of y are determined. 

4. X, the variable to which values are assigned at will is called 
the independent variable, or the argument of the function. 

5. y, whose values are thereby determined, is called the de- 
pendent variable. 

6. y is said to be a function of several variables u, v, w, • • • 
if, when u, v, w, • • • are given, one or more values of y are 
determined. 

7. The variables u, v, w, ' • • , to which values are assigned 
at will are called the independent variables, or the arguments of the 
function. 

Functions of a single variable or argument are represented 
by symbols such as the following: f{x), F{x), 4>{x), yp{x). Func- 
tions of several arguments are represented by symbols such as 
f{u, v, w), F{u, v, w), 0(w, V, w). 

2. The Power Function. $. The function x", where n is a 
constant, is called the power function. 

1 



CALCULUS 



(§2 



If n is positive the function is said to be of the parabolic type, 
and the curve representing such a function is also said to be of the 
parabolic type. If w = 2, the curve, y = x"^, is a parabola. 



-^y 




4-Li'' 



Fig. 1.— Curves for y = x", n = 1, 2, 3, and 4. 

If n is negative the function x" is said to be of the hyperbolic 
type, and the curve representing such a function is also said to be 



§3] 



INTRODUCTION 



of the hyperbolic type. If n = —1, the curve, y = x-^, is an 
equilateral hyperbola. 

In Figs. 1, 2, 3, and 4, curves representing ?/ = x" for different 
values of n are drawn. In Fig. 1, n has positive integral values; 
in Fig. 2, positive fractional values; in Fig. 3, negative integral 
values; and in Fig. 4, negative fractional values. The curves for 



StY" 




Fig. 2. — Curves for y = x", n = 5, 5, f, and |. 



y = x" all pass through the point (1, 1). They also pass through 
the point (0, 0) if n is positive. If n is negative, they do not 
pass through (0, 0). In the latter case the coordinate axes are 
asymptotes to the curves. 

3. The Law of the Power Function. 9. In any power function, 
if X changes by a fixed multiple, y also changes by a fixed multiple. 

The same law can be stated as follows: 



CALCULUS 



m 



10. In any power function, if x increases by a fixed percent, 
y also increases by a fixed percent. 

The preceding statements are also equivalent to the fol- 
lowing: 

11. In any power function, if x runs over the terms of a 
geometrical progression, then y also runs over the terms of a 
geometrical progression. 




Fig. 3. — Curves for y = x", n = —1, —2, and —3. 

4. Polynomials. Algebraic Function. 12. A polynomial in 
a; is a sum of a finite number of terms of the form ax", where a is a 
constant and n is a positive integer or zero. For example: 

ax^ + bx^ -^ ex -\- d. 

13. A polynomial in x and y is a sum of a finite number of 



INTRODUCTION 



terms of the form ox'»i/'», where a is a constant and m and n are 
positive integers or zero. For example: 

axhj^ + ^xy^ + cx^ -{• dy -{• e. 
1 i. Functions of a variable x which are expressed by means of a 
finite number of terms involving only constant integral and 

Y 



w=--|- 







Fig. 4. — Curves for y = x^, n ■= — |, — 5, — f, and — |. 

fractional powers of x and of polynomials in x are included in the 
class of functions known as algebraic Junctions^ of x. For example : 

(a) x\ 



(d) ^, + 1 + 1- 



(b) a;3 + (2a; 



3)i. 



a;'' ■ X 
ie) X + 5 + 



(c) Vx^ + 4x + 7 + 4x + 5. (/) 



■\/x — 7 
3a;2 + 5a; + 7 



x3 _ 3a; + 2 

1 A function of x defined by the equation F(,x, y) = 0, where F{x, y) is a polynomial 
in X and y, is an algebraic function of x. For example, y = \/xs + 2 is an algebraio 
function of x. For by squaring and transposing we obtain 

2/2 - x2 - 2 = 0, 
in which the first member is a polynomial in x and y. 



6 CALCULUS f§5 

15. An algebraic function is said to be rational if it can be 
expressed by means of only integral powers of x together with 
constants. 

Rational algebraic functions are divided into two classes: 
rational integral functions and rational fractional functions. 

16. A rational integral function of x is a polynomial in x. 

17. A rational fractional function is a quotient of two poly- 
nomials in X. 

It is usually desirable to reduce rational fractional functions 
of a: to a form in which the numerator is of lower degree than 
the denominator. This can always be done by performing long 
division. 

X + 3 2 

Thus y = . 1 is equivalent to y = 1 -\ ^rjy, and 

Sx^ + 5x + 7 . . , ,, o , Ux + 1 

5. Transcendental Functions. The circular (or trigonometric), 
the logarithmic, and the exponential functions are included in the 
class of functions known as transcendental^ functions. 

6. Translation. If, in the equation of a curve 

fix, y) = 0, 
X is replaced by (x — a), the resulting equation, 

fix - a, y) = 0, 

represents the first curve translated parallel to the axis of a; a 
distance a; to the right if a is positive; to the left if a is 
negative. 

If y is replaced by (y — /3) the resulting equation, 

fix, 2/ - /3) = 0, 

represents the original curve translated parallel to the axis of y 
a distance j8; up if /3 is positive; down if /3 is negative. Thus 
V = (x + 3)^ — 4 is the parabola y = x- translated three units 
to the left and four units down. See Fig. 5. 

* All functions which are not algebraic functions as defined by the footnote on p. 
6 are transcendental functions. 



§7] 



INTRODUCTION 



7. Elongation or Contraction, or Orthographic Projection, of a 

X 



The substitution of - for x in the equation of any locus 



Locus 

multiplies all of the abscissas by a. 



X-f 




This transformation can be considered as the orthographic 
projection of a curve lying in one plane upon another plane, the 
two planes intersecting in the axis oiy. If a < 1 the second curve 
is the projection of the former curve upon a second plane through 




the F-axis and making an angle a, whose cosine is equal to a, 
with the first plane. If a > 1, the first curve is the projection of 
the second when the cosine of the angle between their planes 

. 1 

IS -. 



CALCULUS 



y 



Similarly the substitution of - for y in the equation of a locus 

multiplies the ordinates by a. The interpretation from the 
standpoint of orthographic projection is evident from what has 
just been said. See Figs. 6 and 7. 

8. Shear. The curve y = f{x) + mx is the curve y = f{x) 
sheared in the line y = mx in such a way that the y-intercepts 
remain unchanged. Every point on the curve y = f{x) to the 
right of the F-axis is moved up (down if m is negative) a distance 
proportional to its abscissa; and every point to the left of the Y- 




axis is moved down (up if m is negative) a distance proportional 
to its abscissa. The factor of proportionality is m. 

In general a curve is changed in shape by shearing it in a line. 
The parabola is an exception to this rule. 

Thus y = dx^ sheared in the line y = mx becomes 

y = ax^ + mx, 
or 

y = ^['' + 2a) -ia' 
This may also be considered as the result of translating the 



§11] 



INTRODUCTION 



9 



4M tfl 

original curve by the amounts — ^ and — j- in the x and ij direc- 
tions, respectively. Hence, by shearing, the parabola y = az^ 
is merely translated. 

9. The Function a*. In Fig. 8 are given the graphs of ?/ = a*, 
for the values a = 1, 2, and 3. By reflecting these curves in the 
line y = X we have the corresponding curves for y = logo x. 

The exponential function y = a" has the property that if x is 
given a series of values in arithmetical progression the corre- 
sponding values of y are in geometrical progression. 




Fig. 8. — Curves for y = a*, a = 1, 2, and 3. 



10. The Function sin x. The function y = sinx is repre- 
sented in Fig. 53. 

11. The Functions p = a cos 9, p = 6 sin ^, and p = a cos 6 + 
b sin 6. The function p = a cos d is the circle OA, Fig. 9, and 
p = bsin d is the circle OB, Fig. 9. The function p = a cos 6 + 
6 sin d can be put in the form p = R cos {8 — a), where 



10 



CALCULUS 



[§12 



R = \/a^ + 6S and where cos « = n and sin a = „ • This function 

is represented by a circle, Fig. 9, passing through the pole, with 
diameter equal to R, and with the angle AOC equal to a. The 
maximum value of the function is R and the minimum value is 
-R. 

12. Fimdamental Transformations of Functions. It is valuable 
to formulate the transformations of simple functions, that most 
commonly occur, in terms of the effect that these transformations 
have upon the graphs of the functions. The following list 
of theorems on loci contains useful facts concerning these 
transformations : 




Fig. 9. 



THEOREMS ON LOCI 

I. If X be replaced by {—x) in any equation containing x and 
y, the new graph is the reflection of the former graph in the F-axis. 

II. If y be replaced by {—y) in any equation containing x and 
y, the new graph is the reflection of the former graph in the X-axis. 

III. If X and y be interchanged in any equation containing 
x and y, the new graph is the reflection of the former graph in the 
line y = X. 

IV. Substituting / - ) for x in the equation of any locus multipUes 
all abscissas l)y a. 



§12] INTRODUCTION 11 

V. Substituting L\ for y in the equation of any locus multiplies 

all ordinates of the curve by b. 

VI. If (x — a) be substituted for x throughout any equation, 
the locus is translated a distance a in the a;-direction. 

VII. If (y — b) be substituted for y in any equation, the locus 
is translated the distance b in the ^/-direction. 

VIII. The addition of the term mx to the right side of 
y = fi^) shears the locus y = f(x) in the line y = mx. 

IX. li {6 — a) he substituted for 9 throughout the polar equa- 
tion of any locus, the curve is rotated about the pole through the 
angle a. 

X. If the equation of any locus is given in rectangular coordi- 
nates, the curve is rotated through the positive angle a by the 
substitutions 

x cos a + y sin a for x 
and 

y cos a — x sin a for y. 

Exercises 
1. Translate the curves 

(a) y = 2x2, (e) y = e", (t) y = -' 



(b) y = - 3X2, (_/■) y = ^3^ (y) y = _. 

(c) y = log X, (g) y = sin x, (/b) y = x^, 

2 

(d) y = e-^, (h) y = cos x, (l) y = x^, 

two units to the right; three units to the left; five units up; one unit 
down; two units to the left and one unit down. Sketch each curve 
in its original and translated position on a sheet of squared paper. 

2. Shear each curve given in Exercise 1 in the line 2/ = ix; 
y = — \x', y = x; y = — x. Sketch each curve in its original and 
shearetl position. 

3. Write the equation of each curve given in Exercise 1 when re- 
flected in the X-axis; in the F-axis; in the line y = x] in the line 
y = — X. Sketch each curve before and after reflection. 

4. Rotate tlie ^curves 



12 CALCULUS [§12 

(a) p = a sin 9, (e) p = a(l + cos $), 

(b) p = a cos 6, (f) p = a(l — sin 6), 

(c) p = a cos ^ + 6 sin ^, (^) p = a(l + sin 0), 

(d) p = a (1 — cos 6), (h) p = ad, 

about the pole through an angle o' 7-' k' ''^j ~ o' *^ketch each curve 

in its original and rotated position. 

6. Sketch the following pairs of curves on squared paper: 

(a) y = x^ and y = x^ -{- x. 

(6) y = x^ and y = (x — Sy + 2. 

(c) y = x^ and y = — a;^ — 2x. 

(d) 2/ = X* and y = a;* — 4a;' + Gx'^ — 4a;. 

(e) y = — 2x'^ and j/ = fx^. 

(f) y = x^ and 2/ = |x^. 
(?) 2/ = sin X and y = sin 2x. 
(h) y = sin x and y = 2 sin x. 

(i) y = cos X and y — sin ( „ — x j . 

6. Rotate the following curves about the origin through the angle 
indicated. 



(o) x* - J/2 = a" 


through 


45°. 


(6) x2 - y2 = a2 


through 


-45°. 


(c) x2 - 7/2 = a2 


through 


90°. 


(d) a;2 - ?/ = a2 


through 


-90°. 


(e) a;2 + 2/2 = a 


through 


a. 


if) y = mx^ 


through 


a. 






CHAPTER I 
DERIVATIVE 

In Elementary Analysis the student investigated the dependence 
of a function upon one or more variables with the help of algebra 
and geometry. 

He is now to study a very powerful method of investigating the 
behavior of functions, the method of the infinitesimal calculus, 
which was discovered by Newton and Leibnitz in the latter part 
of the 17th century. This method has made possible the great 
development of mathematical analysis and of its applications to 
problems in almost every field of science, particularly in engi- 
neering and physics. 

13. Increments. Let us consider the following examples which 
illustrate the principles of the calculus: 

Example 1. A steel bar, subjected to a tension, will stretch, 
and the amount of stretching, or the elongation, will continue to 
increase as the intensity of the force applied increases, until 
rupture occurs. The elongation is a function of the applied force. 
In fact, if the force is not too great, so that the elastic limit is not 
exceeded, experiment has shown that the elongation is propor- 
tional to the applied force (Hooke's Law). If we denote the 
elongation by y and the force by x, the functional relation between 
them will be expressed by the simple equation 

y = kx, 

where A; is a constant. This relation is represented graphically 
by a straight line through the origin, Fig. 10. 

Suppose that after the bar has been stretched to a certain length, 
the force is changed. This change in the force produces a cor- 
responding change in the elongation, an increase if the force is 
increased, a decrease if the force is decreased. Evidently, from 
the law connecting the elongation and the force, this change in the 

13 



14 



CALCULUS 



[§13 



elongation is directly proportional to the change in the force. We 
shall call the change in the force x, the increment of the Jorce, or 
the increment of x, and shall denote it by the symbol A.x (read "in- 
crement of x" or "delta x"). The corresponding change in the 
elongation we call the increment of the elongation, or the increment 
of y, and denote it by Ly. 

In Fig. 10 let P be any point on the line y — kx. If x takes on 
■an increment Ax, y takes on an increment Ay. We see that the 

Ay . 
ratio of these mcrements, i.e., the quotient -r- is entirely inde- 
pendent of the magnitude and sense of Ax and of the position of P 
on the line. Indeed this ratio is the slope of the line. Here the 
increment of y is everywhere k times the increment of x. 




Fig. 10. 



The relation between Ay and Ax can be shown without the use 
of the figure as follows: If x is given the increment Ax, y takes on 
an increment Ay so that 

y + Ay = k{x + Ax). 



On subtracting 






y = kx, 




Ay = kAx. 


Hence 






Ax ' 



a quantity independent of x and of Ax. 

Example 2. A train is moving along a straight track with a 
constant velocity, i.e., it passes over equal distances in equal inter- 



§13] 



DERIVATIVE 



15 



vals of time. Denoting by s the distance measured, say in miles, 
from a fixed point, and by t the time measured, say in hours, 
which has elapsed since the train passed this point, the functional 
relation between s and t is expressed by 

s = ct, 

where c is a constant denoting the velocity of the train. This 
function is represented graphically by a straight line. Fig, 11. If 
we take an increment of time At following an instant t and measure 

As 
the distance As passed over in this time, the quotient r-. repre- 
sents the velocity of the train, since we have assumed the velocity 

As 
of the train to be uniform. Furthermore, the quotient -r^ will 



oo 




>^ 


i 

5 


^^^t 


AS 


^ 




o 


Time ( t ) 





Fig. 11. 

be independent of the length of A^ and of the time t to which the 
increment was given. As is everywhere c times A^. This is 
evident from the graph. 

In these two examples the functions were both linear functions 
of the independent variable. We have seen in these cases (and 
clearly the same is true for any linear function, y = ax -\- h) 
that the ratio of A?/ to Ax is constant. Ay is everywhere equal to a 
constant times Aa;, no matter how large Ax is taken and no matter 
at what point (x, y) on the graph the ratio is computed. 

Example 3. Let us now take an example in which the func- 
tional relation is no longer a linear one. We shall find that the 
ratio of the increment of the function to the increment of the 
variable is no longer constant. Suppose that the train of Example 

As 
2 is not moving with constant velocity. Then the quotient -r--. 



16 



CALCULUS 



[§M 



IS called the average velocity of the train during the interval of 
time At. Evidently this quotient will approximate more and 
more closely to a fixed value the smaller the interval of time 

At, is chosen. The limiting value 

As 
of the quotient -rr as At approaches 

zero is called the velocity at the time t. 
Let the curve of Fig. 12 repre- 
sent graphically the relation be- 
As 
— tween s and t. The ratio t-; cal- 




Tlme (t) 

Fia. 12. 



At 



culated at any point P on the curve 
is no longer constant as in Example 
2, but varies with At and also with the position of the point P. 



the power function 



14. The Fxinction y = x^. Consider 
y = x^. Let us find the ratio of Ay to 
Ax at a certain point of the curve, say 
(0.2, 0.04), for different values of Ax. 
The results are given in the adjoining 
table. 

We observe that as Ax is taken smaller 

Ay 
and smaller the ratio v^ approaches more 

and more closely a value in the vicinity 
of 0.4. 

Ay 
The value of -r— will now be calculated 
Ax 

for any point P, (x, y), on the curve y =x^. 

From this value, which is a function 

of X and Ax, the limiting value as Ax approaches zero will be 

found. The point P, Fig. 13, has the abscissa x. If we give to x 

an increment Ax, we have corresponding to the abscissa, x + Ax, 

the point Q on the curve. Its ordinate is 

y + Ay •= {x -\- Ax)2. 

Ay is equal to the difference between the ordinates of P and Q, or 

Ay— (x + Ax)^ — x^ 
= 2x Ax + (Ax)2. 



Ax 


Ay 


Ay 

Ax 


0.4 


0.32 


0.8 


0.2 


0.12 


0.6 


0.1 


0.05 


0.5 


0.05 


0.0225 


0.45 


0.02 


0.0084 


0.42 


0.01 


0.0041 


0.41 


0.005 0.002025 0.405 


0.002 0.000804 0.402 


0.0010.0004010.401 

\ 1 



§15] 
Then 



DERIVATIVE 



17 



Ay 
Ax 



= 2x + Ax. 



As Ax approaches zero the first term jn the second member remains 

unchanged, while the second term approaches zero. It follows 

• • » ^y 

that the limiting value of v- as Ax approaches zero is 2x. This 

result is expressed by the equation 



I 




Fig. 13. 



Ay 



(read "limit of -r^ as Ax approaches zero"). When x = 0.2, 



limAy 



= 0.4. This is the limiting value which the ratio tabu- 
lated in the last column of the table above is approaching. When 
^ = 3, l^o^x = 6. When x = |, H^.f^ = 1. Thus the 

formula just obtained enables us to calculate very easily, for any 

Ay 
value of X, the limit of -r- as Ax approaches zero. 

16. Slope of the Tangent. — The curve of Fig. 14 is the graph of 
the function y = f(x). On this curve take the point P with 



18 



CALCULUS 



[§16 



coordinates x and y, and a second point Q with coordinates 
X -\- Ax and y + Ay. Draw the secant PQ making tlie angle <^ 
with the X-axis and the tangent PT' making the angle t with the 

Ay 
X-axis. From the figure, -r- is the slope of PQ, or 



Ay 
Ax 



= tan (/>. 



As Ax is taken smaller and smaller the secant PQ revolves about 
the point P, approaching more and more closely as its limiting 
position the tangent PT', and tan <^ approaches tan t. (The 



Y 














/ , 


T 
/ 






J / 






P 


/ 


n 








X 


O 


4 









Fig. 14. 

student will recall that the tangent to a curve at a point P is 
defined as the limiting position of the secant PQ as the point Q 
approaches P.) Hence 



hm Ay 



aj=oAx ^^^=0 ^ 



Hence ^x^qX ^^ equal to the slope of the tangent to the curve 

y = /(^) oi the point for which this limit is computed. 

In the case of the parabola, y = x^, the slope of the tangent at 

2 



§15] 



DERIVATIVE 



19 



the point (x, y) is 2x. This shows that the curve becomes steeper 
and steeper for larger positive and negative values of x and that at 
X = the slope is zero. 

In Fig. 15, let the X-axis be divided uniformly and let the F-axis 
be divided in such a way that distances measured from on a 
uniform scale are equal to the squares of the numbers affixed to 
the points of division. Draw 
lines parallel to the F-axis 
through equidistant points 
on the X-axis and lines par- 
allel to the X-axis through 
points on the F-axis whose 
affixed numbers on the non- 
uniform scale are equal to 
the numbers affixed to the 
points on the X-axis through 
which lines were drawn. 

On the cross section paper 
thus constructed, any point 
at the intersection of a hori- 
zontal and a vertical line 
bearing the same number is 
a point on the curve ij = x^ 
which would be constructed 
in the usual way by using 
the uniform scale on the F- 
axis as well as on the X-axis. 

Join the consecutive points 
thus located by straight lines. 
These lines are the diagonals 
of the rectangles on the cross 
section paper and they are 
secants of the parabola y = x 

let PR = ^x 



Y 




















2 

1 
















/ 




8 












Q 


J 




it 










p 


1 


R 






Y 










1 










3 








/ 












4 
1 






/ 














2 

1 




y 
















4 


_^,,„0^^ 
















„ 







~ 


I - 


^ 


' 


. . 


f - 


r ; 


[ 



Fig. 15. 

Let PQ be such a diagonal and 
RQ Ay 



Then RQ = Ay and ^ = ^. the slope of the 

secant PQ. The diagonals give an approximate idea of the slope 
of the curve. The construction shows why the slope increases 
so rapidly with x. 



20 CALCULUS [§16 

As more and more horizontal and vertical lines are inserted, the 
diagonals approach more and more nearly the direction of the 
tangent lines. 

The fact that the slope of the tangent to the parabola y = x^ 
is 2x furnishes an easy way of constructing the tangent at any 
point P {x, y). We have only to draw from P in the direction of 
the positive X-axis, a line PK of unit length, and from the ex- 
tremity of this line, a line KT parallel to the F-axis, whose length 
is twice the abscissa of P. The line joining PT" is the tangent to the 
parabola at P. When the abscissa is negative the line XT is to be 
drawn downward. 

16. Maxima and Minima. The algebraic sign of ^^q ~ 

enables us to tell at once where the function y is increasing and 
where it is decreasing as x increases. For, if the slope is positive 
at a point, the function is increasing with x at that point and the 
greater the slope the greater the rate of increase. Similarly if the 
slope is negative, the function is decreasing as x increases. Hence 
the function y = x^ is a decreasing function when a;<0 and an 
increasing function when x>0, since the slope is equal to 2x. 
When X = the slope is zero and the tangent is parallel to the 
X-axis. Since the function is decreasing to the left of x = 
and increasing to the right of this line, it follows that the function 
decreases to the value zero when x = and then increases. This 
value zero is a minimum value of the function y = x"^. In general 
we define minimum and maximum values of a function as follows: 

Definition. Let y = f{x), where /(x) is any function of a single 
argument. If y decreases to a value m as x increases and then 
begins to increase, m is called a minimum value of the function. If 
y increases to the value M as x increases and then begins to decrease, 
M is callec a maximum value of the function. 

Thus in Fig. 16, if ABDFHI is the graph oi y = f(x), the func- 
tion increases to the value represented by the ordinate bB and 
then begins to decrease. bB is then a maximum value of the 
function. Similarly fF is another maximum value. dD and hH 
are minimum values of the function. 

In referring to the graph of a function, points corresponding to 
maximum and points corresponding to minimum values of the 



§17] 



DERIVATIVE 



21 



function will be called, respectively, the maximum and minimum 
points of the curve. Thus B and F, Fig. 16, are maximum points 
and D and H are minimum points of the curve. 

Thus, zero is a minimum value oiy == x^ or (0, 0) is a minimum 
point on the curve y = x^. 

It will be noticed that a maximum value, as here defined, is not 
necessarily the largest value of the function, nor is a minimum 
value the smallest value of the function. A maximum value may 
even be less than a minimum value. 




17. Derivative. We see that the limit of the ratio of the incre- 
ment of the function to the increment of the independent variable 
as the latter increment approaches zero, is very useful in studying 
the behavior of the function. This limit is called the derivative of 
the function with respect to the variable. Hence the following 
definition: 

The derivative of a function of a single independent variable 
with respect to that variable is the limit of the ratio of the increment 
of the function to the increment of the variable as the latter increment 
approaches zero. The derivative of a function y with respect to a 

dy 
variable x is denoted by the symbol ^' This symbol will not be 

considered at present as representing the quotient of two quan- 
tities but as a symbol for a single quantity. Later it will be 
interpreted as a quotient. (See §G1.) It is read, "the derivative 
of y with respect to x." The process of finding the derivative 
is called differentiation. 



22 



CALCULUS 



[§18 



18. Velocity of a Falling Body. As a further illustration of the 
application of the derivative let us attempt to find the velocity of 
a falling body at any instant. The law of motion has been experi- 
mentally determined to be 

s = yt\ 

where s is the distance through which the body falls from rest in 
time t. If s is measured in feet and t in 
seconds, the constant g is 32.2 feet per 
second per second, s is plotted as a func- 
tion of the time in Fig. 17. At any time 
t, let t take on an increment A^ s will take 
on an increment As, represented in the 
figure by the line RQ. Since s = \gt^, 







1 


•0 


Ql 








o 






» 






a 






S 




AS 








P 


P/ X 


R 




Ac^t 



Time (t) 

Fig. 17. 



Hence 



s + As = ^git + Aty. 

As = hgit + Aty - hgt\ 



(1) 



or 



As = gtAt + hg{Aty. 



(2) 

This is the distance through which the body falls in the interval 

As 
At counted from the time t. The quotient -r- is the average 

velocity for the interval At. The velocity at t has been de- 

1" As 
fined as ^™ ^^, i.e., as the derivative of s with respect to t. 

To find this limit divide (2) by At and obtain 
As 



At 



= gt + \gAt, 



the average velocity for the interval A^. From which 

lira ^ 



At = At 



= gt, 



or 



ds 
dt 



= gi, 



(3) 



(4) 



the velocity at t. Thus the velocity at the end of three seconds 
is 96.6 feet per second; at the end of four seconds, 128.8 feet per 
second. 



§20] DERIVATIVE 23 

19. Illustration. As an example of the use of the derivative in 
studying the behavior of a function, let us consider the power 
function 

y = x^. 
y + Ay = (x + Axy, 

y + Ay = x^ -\- 3a;2(Ax) + 3x(Aj;)2 + {Axy, 
Ay = 3x2(Aa;) + 3x(Aa;)2 + (Ax)^, 



^ = 3x2 + SxiAx) + (Ax)\ 



Then 



Hm Ay _ 
Ax=0 Ax~ 

T- = 3x2. 
ax 

For X = the derivative is equal to zero and consequently the 
tangent at (0, 0) is horizontal and coincides with the X-axis. 
For all other values of x the derivative is positive. This 
shows that the function is an increasing function for all these 
values of x. Where is the slope of the curve equal to 1? 
Equal to Vs? 

20. Illustration. The solution of the following problem will 
further illustrate the use of the derivative. 

Find the dimensions of the gutter with 

the greatest possible carrying capacity ^ ^ p 

and with rectangular cross section, which 

can be made from strips of tin 30 inches 

wide by bending up the edges to form ^ 

the sides. See Fig. 18. Fig. 18. 

If the depth MR is determined, the 
width is also determined, since the sum of the three sides MR, 
PQ, and RQ is 30 inches. We seek to express the area of the 
cross section as a function of the depth. Denote the depth by 
X and the area by A. The width RQ is 30 — 2x. Hence 

A = (30 - 2x)x. 
In Fig. 19, A is plotted as a function of x. A first increases with x 



24 



CALCULUS 



[§20 



and then decreases. The value of x for which A reaches its 
greatest value can be determined from a graph with a high degree 
of approximation. The derivative can be used to calculate 
accurately this value of x and this saves construction of an 
accurate graph. 

From to H, A is an increasing function. Its derivative is 
therefore positive for this part of the curve . From H to N the 
function A is decreasing. Its derivative is therefore negative 
for this part of the curve. At the point H the derivative changes 
sign, passing from positive values through zero to negative values. 
The abscissa of the point H can then be found by finding the 




Fig. 19. 



derivative of A with respect to x and determining where it changes 
sign. In this case the change of sign occurs where the derivative 
is equal to zero. We find by the method of increments 



ax 



Ax = 4(7.5 - x). 



= when x = 7.5. If a:<7.5, ~r~ is positive and A is an 



dx 



dx 



dA 



increasing function. If x>7.5, -7— is negative and A is a de- 
creasing function. This shows that A increases up to a certain 
value at a; = 7.5 and then begins to decrease. Hence the gutter 



§20] 



DERIVATIVE 



25 



will have the greatest cross section if its depth be made 7.5 
inches. 

It is interesting to plot the derivative as a function of x on the 
same axes. See the dotted line, Fig. 19. The statements made 
concerning the derivative are verified in the graph. 



Exercises 

1. Consider the function y = /(x) whose graph is given in Fig. 20. 
In what portions of the curve is the derivative positive? In what 
portions negative? Where is the derivative equal to zero? 



y =f(x) 




Fig. 20. 



2. Find ,, U y = 3x*. For what values of x is the function in- 
creasing? For what values decreasing? At what point does the 
tangent line drawn to the curve representing the function, make an 
angle of 45° with the positive direction of the axis of x? Find the 
coordinates of the maximum or minimum points on the curve. 

3. Answer questions asked in Exercise 2, if y = x'. 

4. Answer questions asked in Exercise 2, ii y = x*. 

5. Answer questions asked in Exercise 2, if y = x^. 

6. Answer questions asked in Exercise 2, if y = x* — 2x + 3. 

7. Answer questions asked in Exercise 2, ii y = -^ 2" + 2x — 6. 

8. Find the derivative of \/x. 
Solution. Let y = \/x. 

Then 

y +Ay = y/x +Ax 
and 

Ay = y/x + Ax — Vx 
Ay _ \^x + Ax — \/x 
Ax ~ Ax 



26 CALCULUS [§20 

Rationalize the numerator : 

Ay 1 



Ax Vx + Ax + Vx 

As Ax approaches zero the right-hand side of this equation approaches 

-7^- 7^- Then 

Vx + vx 



or 





lira A^ 


1 
2Vx 






1 

2\/x 


9. 


Find the derivative of v/x - 


^. 


lU. 


Find the derivative of \/x* - 


-4. 



CHAPTER II 
LIMITS 

In §17 the derivative was defined as the limit of a certain ratio. 
The word limit was used without giving its precise definition, 
as the reader was supposed to have a fair conception of the mean- 
ing of this term from previous courses in mathematics. How- 
ever, since the entire subject of the calculus is based on limit 
processes it is well to review the precise definition and to state 
certain theorems from the theory of limits. 

21. Definition. // a variable changes by an unlimited number 
of steps in such a way that, after a sufficiently large number of steps, 
the numerical value of the difference between the variable and a 
constant becomes and remains, for all subsequent steps, less than any 
preassigned positive constant, however small, the variable is said to 
approach the constant as a limit, and the constant is called the limit 
of the variable. 



1 



K 



A Xi X2 Xs Xi B 

Fig. 21. 

Illustration 1. Let AB, Fig. 21, be a line two units in length, 
and let x be the distance from A to a point on this line. Suppose 
that X increases from by steps such that any value of x is greater 
than the preceding value by one-half of the difference between 2 and 

2 — X 
this preceding value, i.e., by — ^ — Xi, Xi, Xz, Xi, • • - are the 

end points of the portions of the line representing the successive 
values of x. Then the lengths X\B = 1, x^B = ^, XzB = (^)^ 
XiB = (^)', • • • , XuB = (^)""^ are the successive differ- 
ences between the constant length 2 and the variable length x. 
This difference becomes and remains less than any preassigned 
length KB after a sufficient number of steps has been taken. 

27 



28 



CALCULUS 



[§21 



This is true however small the length KB is chosen. Therefore, 
by the definition of the limit of a variable, 2 is the limit of the 
variable x. 

Illustration 2. Consider the variable x"^ — 2. Give to x the 

3(2" - 1) 



vnliiAQ ft ^ A iJ- A.1 A3. 

values u, 2 , 4 , a i 1 « , 32, 



2" 



= 3[x-i]. 



which are chosen by starting with x = and giving to it successive 
increments which are one-half the difference between 3 and the 



7 


























B 
























8 


6 


























R 





























5 
















































a 






4 






















































«3 


















^ 


, 


































2 


























































































































< 


) 























1 




a 


! 




5 






i 


\ 



Fig. 22. 

preceding value of x. The corresponding values of a;^ — 2 are 
given in the adjoining table. The corresponding points, excepting 
(0, - 2), are plotted in Fig. 22. 

From the table and the expression a;^ — 2 it 
is readily seen that the difference between 7 
and the variable x^ — 2 becomes and remains 
less than any previously assigned quantity (such 
as KB, Fig. 22) after a sufficiently large number 
of steps. Therefore 7 is the limit of the vari- 
able a;2 — 2 as X approaches 3. 

Illustration 3, By giving x values nearer 

and nearer 2, the value of ?: becomes nu- 

' x — 2 

merically larger and larger. Indeed its numeri- 
cal value can be made greater than any preas- 
signed positive number however large by choosing x sufficiently 



X 


x'' -2 


0.0 


-2.00 


1.5 


0.25 


2.25 


3.06 


2.62 


4.89 


2.81 


5.91 


2.91 


6.45 


2.95 


6.72 


2.98 


6.86 


2.99 


6.93 



§23] LIMITS 29 

near 2. The variable does not approach a limit as x ap- 

proaches 2. Instead of doing so it increases without limit. 

// a variable changes by an unlimited number of steps in such 
a way that after a sufficiently large number of steps its numerical 
value becomes and remains, for all subsequent steps, greater than 
any preassigned positive number however large, the variable is said 
to become infinite. Illustration 3 of this section is an example 
of a variable which becomes infinite, or approaches infinity. 

22. Notation. If in any limit process, the variable, say y, is a 
function of another variable, say x, the successive steps by 
which y changes are determined by those by which x changes. 
// y approaches a limit A, as x approaches a limit a, we say that the 
limit of y as X approaches a is A, and write 

lim ., _ 



x=a 



y = A. 



After what has just been said, the meaning of the two following 
expressions will be clear: 

lim y ^ ^ lim„ = ^o. 

In the second case a limit does not really exist. The form of 
expression is only a convenient way of saying that if x is taken 
sufficiently near a, the value of y can be made to become and 
remain greater in numerical value than any preassigned positive 
number however large. 

From the illustrations of the preceding section we have: 

1. ""^ X = 2, where n is the number of steps taken. 

2. ^i"^ 

z=3 

Q lim 



X.3 (^^ - 2) 



"2= "'^ 



23. Infinitesimal. In the particular case where the limit of a 

variable is zero, the variable is said to be an infinitesimal. An 
infinitesimal is a variable whose limit is zero. Thus Ay and Ax 
which were used in §§13, 14, and 15 are thought of as approaching 



30 CALCULUS [§24 

zero and are infinitesimals. Hence the derivative, §17, is defined 
as the limit of the quotient of two infinitesimals. Infinitesimals 
are of fundamental importance in the Calculus. Indeed the 
subject is often called the Infinitesimal Calculus. 

24. Theorems on Limits. The following theorems concerning 
limits are stated without proof: 

Theorem I. 7/ two variables are always equal and if one approaches 
a limit, the other approaches the same limit. 

Theorem II. The limit of the sum of tivo variables, each of which 
approaches a limit, is equal to the sum of their limits. 

Theorem III. The limit of the difference of two variables, each 
of which approaches a limit, is equal to the difference of their limits. 

Theorem IV. The limit of the product of two variables, each of 
which approaches a limit, is equal to the product of their limits. 

Theorem V. The limit of the quotient of two variables, each of 
which approaches a limit, is equal to the quotient of their limits, 
provided the limit of the divisor is not zero. 

If the limit of the divisor is zero, the quotient of the limits in 
Theorem V has no meaning, since division by zero is an impossible 
operation. For, the quotient Q of two numbers A and B is 
defined as the number such that when it is multiplied by the 
divisor B, the product is the dividend A. Now if B is zero while A 
is not zero, there clearly is no such number. 

25. The Indeterminate Form 5. If, in the quotient considered 

above, A is also zero, any number will satisfy the requirement, 

so that Q is not determined. One encounters exactly this diffi- 

a;2 _ 4 
culty in seeking the value of ^ at a; = 2. Its value is not 

determined at this point but it is determined for all finite values 
of X different from 2. We define its value at a; = 2 as the limit of 
its value as x approaches 2. The student should construct a 
graph of this function. Usually we proceed as follows to find the 
desired limit. 

lim a;2 - 4 Hm 



x=2 a; _ 2 x=2 



(x + 2) = 4. 



a;2 _ 4 

The expression -_-^ is said to be indeterminate at x = 2, 



§26] 



LIMITS 



31 



since any one of an infinite number of values can be assigned to it. 
The determination of its limiting value as x approaches the value 2 
is called the evaluation of the indeterminate form. Indeterminate 
forms of this and other types are frequently found in the Calculus. 



Thus V- is an indeterminate form for Ax 



0. We have already 

seen in several cases how it can be evaluated. Exactly as in the 
example just given, we have sought the limit of the quotient as 
Ax approaches zero and not the quotient when Ax = 0, because the 
latter quotient has no meaning. 



Exercises 

1. Determine the following limits, if they exist. 



(6) ^^gCOtX. 



, lim 

(a) ^cosx. 

(c) \"} sin -. Draw the curve for values of x between — tt and -f- jr. 

(d) ".'V,xsin"- 

2. Evaluate the following indeterminate forms : 



(a) 



x2 -9 



X -3 
3. Find 



(&) 



X* + 6x2| 



symbol 



z=s=3 ' ' 3x3 + x^\x^O 

lim 3x2 _ jjj^ 4^_ jjj^ 4^2^ jj^^ 4x^ + 3 

j;=oo a; » x=oo53.2> a;===co5^2> x= «= Qx^ ' 

Show that it is an indeterminate form. 



26, Continuous and Discontinuous Functions, 
graphs of the following functions: 

1 



1. y = 



2. y = x^. 



3. y = 7x + 



Discuss the 

Draw the 
1 



4. 7/ = tan X. 5. y = sin x. 6- y = 3^. 

Hint. In 6, values in the vicinity of a; = should be carefully 
determined. Take a set of values of x approaching from the 
left and another set approaching it from the right. 



7. 2/ = 3^ 



3'' + 2 
3' + 1 



32 CALCULUS [§26 

Study the vicinity of a; = 0. See 6. 

The functions 2, 5, and 7 are said to be continuous while 
1, 3, 4, 6, and 8 are discontinuous. The meaning of these 
terms is obvious from the graphs that have been drawn. A 
precise definition follows : A function f{x) is said to be continuous 

atx = a if Ji^ fix) = /(a). 

In 1, 3, 4, 6, and 8, this condition is not satisfied at a; = 0, 

0, ^, 0, and 0, respectively. In these examples the functions 

either become infinite for the values of x in question or approach 
different limits as the value of x is approached from larger or 
smaller values. A function f{x) is said to be continuous in an in- 
terval (c, d), i.e., the interval c ^ x ^ d, if it is continuous at every 
point in this interval. Thus the functions 2, 5, and 7 are continu- 
ous in any finite interval. The remaining functions are continu- 
ous in any interval not containing the points to which attention 
has been called. 



CHAPTER III 
THE POWER FUNCTION 

27. In Chapter I the derivative of a function was found by what 
may be called the fundamental method, viz., by giving to the 
independent variable an increment, calculating the corresponding 
increment of the dependent variable, and finding the limit of the 
ratio of these increments as the increment of the independent 
variable approaches zero. This method is laborious and since it 
will be necessary to find derivatives in a large number of problems, 
rules will be established by means of which the derivatives of 
certain functions can be written down at once. The process of 
finding the derivative of a function is called differentiation. 

In this chapter we shall find the derivative of the power 
function, and study the function by means of this derivative. 

The graphs oi y = x", for various values of n, appear in Figs. 
1, 2, 3, and 4. If n is positive, the curves go through the points 
(0, 0) and (1, 1), and are said to be of the parabolic type. In this 
case a;" is an increasing function of x in the first quadrant. If n 
is negative, the curves go through the point (1, 1) but do not go 
through the point (0, 0). They are asymptotic to both axes of 
coordinates. These curves are said to be of the hyperbolic type. 
In this case x^ is a decreasing function of x in the first quadrant. 

The law of the power function, as stated in §3, should be 
reviewed at this point. 

28. Derivative of x°. Let y = x", where n is at first assumed 
to be a positive integer. 

y -{- Ay = (x + A.r)". 

y+ Ay = X- +nx--''Ax + "^^,"~-"^^x"-2 (Ax)^ + ■ +(Ax)". 

Ay = nx--^Ax + ^^^ a-"-^ {AxY+ ■ ■ +(Ax)". 
If 
a 33 



34 CALCULUS [§28 



Ay , , n{n - 1) 



Ax 
lim Ay 

or 

dy 
dx 



+ .^ a;»-2 Aa;+ • • • +(Ax)»-i. 



= nx"" 



= nx"-^ (1) 



This proof holds when n is a positive integer. In §§33 and 42 
it will be shown that the formula obtained holds for fractional and 
negative exponents. For the present we shall assume the 
formula true for these exponents. 

Illustrations. 

1. ^ = 3.=. 
ax 

dx dx X* 

3. ^' . W. 
dt 



dv ~ ""^ ~ 2\/y 



Exercises 

dy . . . 

Find -y- in each of the following fifteen exercises: 

1. y = x^. 6. y = x. 11. y = x*. 

2. y = x<. 7. y = x*. 12. y = x^. 

3. y = x^ 8. y = x». 18. y = x"*. 

3 _1 

4. y = x^". 9. y = x^. 14. y = x ^. 
6. y = -• 10. y = -^^ 15. y = -,- 

16. Find the slope of each of the curves of Exercises 1-15 at the 
point (1, 1); also at the point whose abscissa is ^ and whose ordinate 
is positive. 

17. By making use of the derivative, find for what values of x 
each function given in Exercises 1-15 is increasing; is decreasing. 



§29] THE POWER FUNCTION 35 

18. How does the slope of j/ = a;" change with increasing x, if x 
is positive and if n is positive and less than 1? If n is positive and 
greater than 1 ? 

19. Find where the slope of each curve given in Exercises 1-15 is 
equal to zero; equal to 1. 



20. Find:riif: 
at 






(a) s = t\ 


(C) 8 = Vt' 


(e) s = ^yr^ 


(b) s=l,- 


(d) s = \- 
t^ 


CO . = ^-- 


21. Find ^f if: 
at 






(a) y = IK 


(c) y = ^l 


(e) y = IK 


ib)y=l 


(d) y = 4- 


U)y=l- 



29. The Derivative of ax°. In case the power function is writ- 
ten in the more general form ax", it is easy to see that the con- 
stant multiplier a will appear as a coefficient in all terms on the 
right-hand side of the equations in the proof in §28, and the 
derivative oi y = ax" is, therefore, 

f^ = nax'^-^, (1) 



or 

d(ax'') 
dx 



— nax"-^. (2) 



The proof of the formula is for positive integral values of n only, 
but as in §28 will be assumed for all commensurable exponents.^ 

Since ax"-^ is the given power function y = ax" divided by x, 
formula (1) may be written 

' The relation of formula (2) to that of §28 is at once evident when it is recalled that 
the curve J/ = oi" can be thought of as obtained from the curve j/ = x" by stretching 
all ordinates in the ratio 1 : a. Then the slope of the tangent at a point of j/ = ox" is 
o times the slope of the tangent to y = i" at the corresponding point; i.e., 

d(ax") d(x'^) , 

= o = anx" . 

dx dx 



36 



CALCULUS 



[§29 



The geometrical meaning of formula (3) is shown by Fig. 23. 

y 

The fraction - is the slope of the radius vector OP from the origin 

to the point P on the curve. Formula (3) states that the slope, 
at any point of the graph of the function, y = ax", is n times the 
slope of the radius vector OP. Thus, if n = 1, ?/ = ax" reduces 
to a straight line through the origin, and the line has the same 
slope as OP. If n = 2 the curve is the parabola y = ax^, and the 
slope of the curve is always twice that of OP. If n = —1 the 



r 


F 


ft 1 

/ 




o 















Fig. 23. 

curve is the rectangular hyperbola, y = 

curve is the negative of the slope of OP. 
Illustrations. 
d{7x^) 



1. 



dx 

il 

dx 



= 14a;. 



d{5x-^) ^ d(x-^) 



and the slope of the 



dx 



= 5 



dx 



= -lOx-3 = - 



10 



3. -^-iT-- = 6 ^TT- = 9^2 



dt 



4. If 2/ = 5x', 



dt 



dy 
dx 



§30] THE POWER FUNCTION 37 

5. If V = ~2' ^~ = -2 



x^ dx X 



Exercises 
Find -T~ in each of the following fifteen exercises . 

1. y = 4x3. 6. 2/ = 2xi 11. y = -3x«. 

2. J/ = 3\/x. 7. 2/ = 4x^. 12. ?/ = x. 

3. y = 5x*. 8. y = fx"". 13. y = -x. 

4. y = 3x. 9. 2/ = lOv^i. 14. 2/ = -3\/x. 

2 4 4 

6. 2/ = -,' 10. 2/ = --2- 15. 2/ = --3- 

16. Find -yj in each of the following: 

(a) s == 2<2. (6) s = SVi". (c) s = -^tK 

civ 

17. Find -jt in each of the following: 

(a) 2/ = 4v^. {b) y = - At. (c) ?/ = -Si^. 

18. Find the slope of each of the curves given in Exercises 1-15, 
at the point whose abscissa is 1; at the point whose abscissa is |. 

19. For what values of x is each function given in Exercises 1-15 

increasing? Decreasing? Where, if at all, is the slope of each of 

these curves zero? 

2 

20. Draw the curves y = \x^, y = ~, y = x^, y = -y/x] and draw 

tangent lines to them at the points for which the abscissas are 1, 2, 3, 
and 4. Make a table showing the slope of the radius vector and the 
tangent line for each of these points. 

30. Rate of Change of ax°. Let y = ax", where a; is a function 
of the time t. Since a: is a function of t,y is a, function of t. For 
example, y = Sx^, where x = t — 1. 

Let Ax and Ay be the increments of x and y, respectively, corre- 

Aw 
spending to the increment A^ of t. -rr is the average rate of 

dv 
change of y during the interval AL .r is the rate of change of y 

at the instant t. 
At any time, t 

y = ax". 



38 CALCULUS [§30 

At the time t + Af, 

y -\- Ay = a{x + Ax)". 

y -\-Ay= alx-+nx"-'Ax + Z ^ x»-''(Ax)H h (Ax)"l • 

Ay = aLx"-' + "^^^^ a;»-2Aa; + • • • + (A.t)"-i1 Ax. 

Aw r , > W(W "~ 1) OA ■ I /A N iTAX 

^^ = a wx»-i + ^2"^ x"-2Ax + • • ■ + (Ax)»-i ^• 

As Af approaches zero, the expression within the brackets ap- 
proaches nx""^, and irr approaches -jr* 

Hence 

dy dx 



dt -°^^""'dr 

or 

. dt "*^ dt 

The rate of change of the function ax" is expressed in terms of x 

dx 
and of -TT, the rate of change of x. If then the rate of change of 

X for a given value of x is known, the rate of change of the func- 
tion for that value of x can be calculated. 

Illustration 1. The side of a square is increasing at the uni- 
form rate of 0.2 inch per second. Find the rate at which the area 
is increasing when the side is 10 inches long. 

Let X be the length of the side, and y the area of the square. 

dx dv 

Then ^rr = 0.2 and ^r is the rate of increase of the area. To find 
dt dt 

this rate of increase, differentiate the function y = x^. 

^ -2x— • 
dt ^^ dt 

Since 

dx 

dt = 0-2. 

dt "-^ 



§31] THE POWER FUNCTION 39 

dv 
When X = 10, -rr = 4. The area is increasing at the rate of 4 

dv 
square inches per second. When x = 13, -tt = 5.2, the rate of 

change of the area at this instant. 

Illustration 2. A spherical soap bubble is being inflated at the 

rate of 0.2 cubic inch per second. Find the rate at which the 

radius is increasing when it is 1.5 inches long. 

dV 
Let r be the radius, and V the volume of the bubble, -rr = 0.2 

at 

dv 
and ^7» the rate of increase of the radius, is to be found. 

V = AirrK 





^ = ^^^^41- 


From which 






dr 1 dV 

dt ^ 47rr2 dt 



Since 

dV 

TT- = 0.2, and r = 1.5, 

dr 1 

tit ~ on-n K \2 ~ 0.0071 inch per second. 

Exercises 

1. Find the rate at which the surface of the soap bubble of Illustra- 
tion 2 is increasing when r = 1.5 inches. 

2. If each side of an equilateral triangle is increasing at the rate 
of 0.3 inch per minute, at what rate is the area of the triangle increas- 
ing when the side is 6 inches long? 

3. Water is flowing at a uniform rate of 10 cubic inches per minute 
into a right circular cone whose semi-vertical angle is 45°, whose apex 
is down, and whose axis is vertical. At what rate is the surface of 
the water in the cone rising, and at what rate is the area of this surface 
increasing when the water in the cone is 25 inches deep? 

31. The Derivative of the Sum of a Function and a Constant. 

Sketch, on the same set of axes, the graphs of the functions: 



40 CALCULUS [§31 



y = x^; y = x^ — 5; y = x^ -{- 3; y = x^ -\- 10. Find -v- for each 

dy 
of the functions. Find t-, if y = x"^ -{- C, where C is any 

constant. 

Sketch a graph of any function y = f{x), and on the same set 
of axes, graphs oi y = fix) -\- C for several values of the constant 

dy 
C. What relation exists between t- for the different functions 

dx 

corresponding to the same value of cc? 

From these illustrations it is clear that the derivatives of all 
Junctions which differ only by an additive constant are the same. 
The reason for this is geometrically evident. For, the addition 
of a constant to a function has the effect of merely translating the 
graph of the function parallel to the F-axis. The slope corre- 
sponding to any given abscissa is clearly not changed by this 
translation. Hence, 

d[f(x)+C] d[f^x)l 



dx dx 

In particular d[axn-|-C] d[ax'i] 

_ = _- — = a 

dx dx 

Thus, if _ , . „ 

dy d{5x^) 



rll— 1 



(1) 

(2) 



dx dx 

Exercises 



15x2 



1. Prove formula (2) above by the increment process. 
Differentiate : 

2. y = 3x^ + 2. 11. 7/ = - 4x^ + 6. 

3. y = 5\/x + 4. 12. y = - 3x* + 2. 

4. y = 2x3 _ 3. 13. y = Tx^ - 3. 

3 

5. 2/ = — 5 + 5. 14. ?/ = 4x' + 5. 

6. 2/ = 2t* + 7. 15. 2/ = ^ + 2. 

7. s = 16<2 + 5. 16. 2/ = - ix« + 3. 

8. « = 2y/t^ + 6. 17. 2/ = ix* + 2. 

9. s = ^3 - 4. 18. 2/ = ?x* - 5. 
10. X = 4<3 - 2. 



§32] THE POWER FUNCTION 41 

32. The Derivative of au°. li y = aw, where w is a function 
of X and n is a positive integer, the student will prove, as in §30, 

that 

dy .du 

dx dx 

or 

d(au'') „ .du 

dx dx 

Illustrations. 

= 30x(x2 + 3)2. 

2 d[2(x'' + 4)^+10] _rf[2(x2 + 4)^]_,^ ,(J(x2 + 4) 

dx dx dx 

= 2-4(a;2 + 4)3 2.r = 16x(x2 + 4)'. 

3. If y = (2x2 + 1)2, find ^. 

dy _ cj(2x 2+l)2 _ , 1^ ^(2x2+ 1) 



•2'2x 
4. If 77 = (x2 + l)t. 



= 2(2x2 + i).2-2x ^ = 8x(2x2 + 1) ^j- 





dy _ 3 
dx ~ " 


(a:2+l)^ 


2x = 


3x(x2 + 1)5, 




and 


dy 3 


(X2+1)^ 


dt 


= 3x(x2 


+ 1)^ 


dx 
dt' 








Exercises 






Fi 


nd^: 
ax 

y = (4x2 

y = 5(2x 

y = 2(3 
y = \/x 
y = (5- 


-2)3. 
2 - 5)3. 

- 4X2)3 

- X2)3. 




6. y 

7. y 

8. y 

9. y 

10. y 






1. 
2. 
3. 
4. 
6. 


= V9 
= (3x2 
= a/x^ 
= V2a 


-X2. 

+ 7)2. 
-5. 

;2 + 3 
;2 + l. 



42 CALCULUS [§33 



11- y = /^ ' • 15. y = 



VSx - 2 A^x2 + 1 

7 "i 

12. 2/ = -7===-. 16. y = 



2 



13. y = V(x« + 4). 17. y = y- 



14. 2/ = V(x + 1). 18. 2/ = 



(a:2 + 4)2 
3 



(5 -x»)2 

33. The Derivative of u*^, n a Positive Fraction. We are now 

in a position to prove that the rule for the derivative of m" holds 

when n is a positive fraction of the form -' where p and q are 

integers. Let 

p 
y = w. 

Raise each member to the power q: 

?/« = UP. 

Since m is a function of x,yisa function of x. Hence each member 
is a function of x raised to a positive integral power. Then each 
member can be differentiated by the rule of §32 which was 
proved for positive integral exponents. We find 









, dy 

^y-' Tx = 


= pu 


-1 '^'^. 
dx 




From which 






dy V 


tip-i 


'du 










dx q y-^ 


dx 




Substitute for y in 


the second member and obtain 




dy 


P 


wp-^ 


du 


p up- 


-1 du 




dx ~ 


9 


u' 


dx 


q 

up- 


p dx 


Then 






dy ^1 
dx q 


u 


du 
dx 





§35] THE POWER FUNCTION 43 

and the rule is proved that 

dw , du 

ax dx 

where n is a positive fraction whose numerator and denominator 
are integers. This rule has already been used in the solution of 
numerous exercises. 

34. The Derivative of a Constant. Let y = c, where c is a 
constant. Corresponding to any Ax, Ay = 0, and consequently 

Ax "' 



lim -^ = 0. 



and 

Ax = Ax 

or 

^ = 0. 
dx 

The derivative of a constant is zero. 
Interpret this result geometrically. 
35. The Derivative of the Sum of Two Functions. Let 

y = u + v, 

where u and v are functions of x. Let Au, Av, and Ay be the incre- 
ments of u, V, and y, respectively, corresponding to the increment 
Ax. 

y -\- Ay = u -{- Au + V -jr Av 

Ay = Au -\- Av 

Ay _ Au Av 

Ax ~ Ax Ax 

dy du . dv 

dx dx dx 
or 

d(u4-v) _ du dv 

dx ~ dx dx* 

The derivative of the sum of two functions is equoiL to the sum of 
their derivatives. 



44 CALCULUS [§36 

The student will observe that the proof given can be extended 
to the sum of three, four, or any finite number of functions. 
Illustrations. 

I t^(63^ + 15a;') ^ d(6x) rf(15x') _ g , 30^ 
dz dx dx 

^ d{2V x + 3x' + 4) d{2V~x) , d(3x^) , d(4) 1 

2. = h. ' = — 7= + ox. 

rfx dz dx dx y/x 

d(t^ + 2<' + 3) 

3. f; = 2< + 6«2 

Exercises 

Differentiate the following functions with respect to x, also with 
respect to t: 

1. 3x* - 2x^ + 6. 10. ax2 + bx + c. 

2. 5x» - 7x2 -2x - 10. 11. y = v^a;2 + 4^; _ 5. 

3. gx* - ix' + X - 7. 12. 2/ = 1 



4. x2 + 2x 
3 2 



y/x^ - 5x + 7 



13. 2/ = V3x2 _ 2x + 5. 

* \7x ~ ^" ^ 14. y = Ve - 3x - x\ 

6. 3x7 _ 63.8 + 9. 16. s = Vm + ^2< - 3 

?_2 1 
.. -. — X ^ 16. 7/ = 

x" - 7x - 6 

8. - \x^ -\- \x^ - X + 2. _ / 

II 17. y = V a;^ - 5x + 4. 

9. x~5 4- a;~5. 18_ y _ (3_j.2 _|. 2x + 2)3. 

36. Differentiation of Implicit Functions. The derivative of 
one variable with respect to another can be found from an equa- 
tion connecting the variables without solving the equation for 
either variable. For, if the variables are x and t/, j/ is a function 
of x, even though its explicit form may not be known, and the 
usual rules for finding the derivative of functions can be applied 
to each member of the equation. 

The following example will illustrate the process. 

dy 
Illustration. Letx' + 2/^ = O"- Find -? • 

The left-hand member of the given equation is the sum of 



§36] 



THE POWER FUNCTION 



45 



two functions of x, since y is a function of x. Further, the deriva- 
tive of the left-hand member is equal to the derivative of the 
right-hand member. The derivative of the latter is in this case 
zero, since the right-hand member is constant. On differentiat- 




x+ y=^o, 



Fig. 24. 



ing the left-hand member as the sum of two functions, we obtain 



Solving for -t-> 



2x + 2y% = 0. 



dx 



When the derivative is found by differentiating each member of 
an equation in the implicit form, as in the foregoing illustration, 
the operation is called implicit differentiation. 



46 CALCULUS [537 

Exercises 

1. Draw the circle x^ + y'' = o* and show geometrically that the 
slope of the tangent at the point (x, y) is 

dy 

2. Solve the equation of Exercise 1 for y and find -j- • 

dv 
From the following equations find -r by implicit differentiation: 

3. 3x« + 4y* = 12. 

4. a;^ — ?/* = a^. 

^' ^ + ^ = !• (^o ^'^^ clear of fractions.) 

If y is an implicit function of x expressed by an equation of the 
form 

x" + y = a", (1) 

differentiation gives 

or 

g=-[i]""' <^> 

The equation (1) includes a number of important special cases. 
The graphs corresponding to the following values of n are shown in 
Fig. 24. For 

n = h x' + y* = o^, a parabola, 

2 1 2 . ■ 

n — I, x^ -\- y^ = a^, an important hypocycloid, 

n = l, X + y =a, a straight line, 

n = 2, x"^ -\- y"^ = o^, a circle. 

The graph of (1) passes through the points (0, a) and (a, 0) if n is 
positive. 

12? 

6. x» + J/* = o^. 

Ill 

7. x* + 2/2 = o2. 

8. x' + y' = a\ 

3 3 3 

9. x^ + y^ = a^. 

37. Anti-derivatives. Integration. Let it be required to 
find the equation of a curve whose slope at any point is equal to 
twice the abscissa of that point. 



§37] THE POWER FUNCTION 47 

dy 
This means that at every point of the curve -, = 2x. We 

seek then a function whose derivative is 2x. y = ^^ is such a 
function. But y = x^ -\- C, where C is a constant, is also a 
function having the same derivative. Hence there is an infinite 
number of functions whose derivatives are all equal to 2x. The 
problem as proposed has then an infinite number of solutions, 
viz., the system of parabolas y = x^ + C, corresponding to the 
infinitely many values of C. 

If now we add to the statement of the problem the requirement 
that the curve shall pass through a given point, say (1, 2), it is 
geometrically evident that but one of the curves y = x^ + C will 
pass through the point. In other words there is but one value of C 
for which the latter requirement is satisfied. This value is de- 
termined by substituting the coordinates of the point in the 
equation y = x^ -{- C, since they must satisfy this equation for 
some value of C, if the problem has a solution. On making the 
substitution we have 

2 = 1 + C, 

from which (7=1. Hence y = x"^ -\- 1 is the equation of the 
curve whose slope at any point is equal to twice the abscissa of the 
point and which passes through the point (1, 2). 

The nature of the problem which has just been solved can be 
further explained by the following geometrical solution. Draw, 
Fig. 25, at the vertices of each small square on a sheet of co- 
ordinate paper on which a set of axes has been chosen, short 
lines whose slopes are equal to two times the abscissas of the 
respective vertices. A curve is to be drawn which at each of its 
points is tangent to a line such as those which have been drawn. 
Now it is impossible in the figure to draw lines through every point 
in the plane, but if the points through which the lines are drawn 
are sufficiently thick, the lines will serve to indicate the direction 
which the curve takes at nearby points. The lines may be 
regarded as pointers indicating the stream lines in flowing water. 
Then a point tracing the curve would move as a small cork would in 
water having the stream lines indicated by the figure. 

Thus, to get the curve that goes through (1, 2), start from this 
point and, guided by the direction lines, sketch in as accurately 



48 



CALCULUS 



I§37 



as possible the curve to the right of this point. Do the same 
thing to the left, noting that here it is necessary to go against the 
stream lines instead of with them. 

In Fig. 25 it should be noted that all lines through points 
having the same abscissa are parallel. This fact is of great 



^ ^ ^ ^ ^ \ 



^ 






M ^ ^ H ^ 






k k h \ \ ^ 
k k k \^\^ 




/MM 



/^ ^ M M 



Fig. 25. 

assistance in drawing. The squares on the coordinate paper can 
be used to advantage in drawing lines when the slope is known. 

If the derivative which was given had been any other function 
of X, a geometrical solution could have been obtained by the same 
method. - 

The foregoing illustration introduces a new type of problem, 



§37] THE POWER FUNCTION 49 

viz., that of finding a function whose derivative is given. A 
function whose derivative is equal to a given function is called 
an anti-derivative, or integral, of the given function. From the 
illustration it is clear that any given function which has one anti- 
derivative, has an infinite number of anti-derivatives which differ 
from each other only by an additive constant. This latter fact is 
indicated in obtaining the anti-derivative of a given function by 
writing down the variable part of the anti-derivative and adding 
to it a c onstant C which is undetermined or ' ' arbitrary. ' ' In a given 
application this constant will be determined by supplementary 
conditions as in the illustration at the beginning of this section. 
The process of finding the anti-derivative of a given function is 
called integration. 



y = x^ + C. 

y = lx' + C. 

y — x^ ■{- x^ -\- C. 

y = x^ + x"^ -\- Ix -\- C. 

2/ = 3- + 2- + 7a; + C. 

If in Illustration 1 the curve is to pass through the point 
(3, -2) we must have -2 = 3' + C, or C = -29. Hence the 
equation of the curve is y = x^ — 29. 

Exercises 



Illustrations. 




1. 


jfdy 
dx ~ 


3x2 




2. 


If — - 
dx 


x\ 




3. 


-ffdy 


3x2 


+ 2x, 


4. 


If — - 
dx 


3x2 


+ 2x -h 7, 


5. 


Tf ^ - 
dx 


x^ + x + r. 



Integrate the following ten functions: 




1- 1 = 3,. 


^- dt ~ 


dx 

(3x2 + 2x + 6) ^ 


.. g = 4... 


^ dy _ 
'' dt 


(ax + b) ^-. 


''•|--4r 


^- dx~ 


3x2 _ 2x2 _|. 7. 


«• f = 3..f . 


Q '^y 

^' dx- 


lOx-2 + 2x-3 - X 


6. ^ = 3x2 + 2x - 6. 


10. f~ = 

dx 


x^ + x'\ 



60 CALCULUS l§37 

Illustrations. 

6. ^ = 3(x2 + 2)2 2x. 

The right-hand side is in the form, nw-^ ^, where n is 3, and w 

du -, 

is (x'^ + 2). Since the integral of nw"-i ^ is -u" + C, 

2/ = (x^ + 2)» + C. 

7. ^ = (x2-5)'2x = H4(x2-5)'2x]. 

2/ = i(x2 - 5)^ + C. 

8. ^ = x(x2- l)'' = -iM6(x2- l)''2x]. 

y = Mx^ - 1)« + C. 

y = _ -j^(3 - x')« + (7. 
10. I = (.' - 2x + 3)-'(x - l);j^ 

= - il - 2(x^ - 2x + 3)-^(2x - 2)^J. 

y " 4(x2 - 2x + 3)2 + ^* 

Exercises 
Integrate : 

11. ^ = a;V^^"^=l- Ans. y = K^^ - 1)^ + C. 

12. ^ = (2x3 4. 3x2) 3 (x^ _|_ x). Ans. y = i(2x3 4. 3x2)1 + c, 
ax 



13. ^ = (x + 1)*. ^ns- y = f (x + 1)^ + C. 



dx 



14. I = (2- XT X. 16. ^^ MX- 3)2 X. 

16.| = xVr^'. 17.g = x2+3x. 

18.^^ = (x2 + 7)3x. 



§37] THE POWER FUNCTION 51 

20. f^ = (x^ + 2x + iy(x + 1). 



22. 


dy 

dx ~ 




X 






24 


dy 
dx 


X2 


V5 


-X* 


- (4 - x")* 


25. 


dy 

dx ~ 


(3x2 


+ 2) 


' X. Ans. 


y 


= A 


(3x2 


+ 2y + c. 


26. 


dy 

dt ~ 


(2- 


- 3x2) 


. dx 
'"'dt- 










27. 


dy ^ 
dx 


(2x 


+ 1)^ 


Ans. y 


= 


K2x 


+ 1] 


3 + C. 


28. 


dy 
dt ~ 


(3x 


-2y 


dx 
dt' 











29.^f = (3-4x)2^. 

30. ^ = V^TT- ^ns. 2/ = f (x + 1)' + C. 

31. ^ = >/2F+^. 40. ^ = \/(2x+3)». 

33. -2 = Vir=^. 42. ^ = Vx^ + 3x - 7 (x^ + 1). 

3*- 5x = vTTT' "• ^' ^ ^" "^ 4)^/iM^8^+^. 

^^- ^^ = virr2 '*• ^' = ^^ - ^)^^^^^+^- 

^^ dy 1 .- dy 1 — 5x 

36. T^ = , 46. " 



dx V3 - 5x * dx -y/e 4- 4a; - lOx* 

— dw / .^ dy 3x — 2 

37. / = X V4x2 _ 5. 46. *' 



dx * ' dx V3x2 _ 4a; _|_ 5 

88. I - .V9^'. «. I - V4^. 

89. f . '^ 48. f^ - x(2 - x>).. 

dx .^^4 — 3x2 dx 

49. Find the equation of the curve whose slope at any point is equal 
to the square of the abscissa of that point and which passes through 
the point (2, 3). 

60. Find the equation of the curve whose slope at any point is equal 



52 CALCULUS I §38 

to the square root of the abscissa of that point and which passes 
through the point (2, 4). 

51. Find the equation of the curve whose slope at any point is equal 
to the negative reciprocal of the square of the abscissa of that point 
and which passes through the point (1, 1). 

38. Acceleration. The velocity of a body moving in a straight 
line may be either uniform or it may vary from instant to instant. 
In the latter case its motion is said to be accelerated, and this 
applies both to the case where there is an increase in velocity 
and the case where there is a decrease in velocity. 

Thus it is a fact of common knowledge that the velocity of a 
body falling to the ground from a height increases with the dis- 
tance through which the body has fallen, or with the time since 
the body started to fall. The time rate of change of the velocity 
of a moving body is an important concept in mechanics and 
physics. 

If s denotes the distance passed over in time t, the velocity has 
been defined as the rate of change of s with respect to t. The 
notion of the velocity at a given instant was derived from that of 
the average velocity for an interval A^ The average velocity was 
obtained by dividing the change in s, As, in a time A^ by At {i.e., 
by dividing the distance passed over in time At by At). The 
limiting value of this quotient as At approaches zero was defined 
as the velocity at the beginning of the interval A^. 

In the same way if the velocity, r, changes by an amount Av in 
the time A^ counted from a certain time t, the average rate of 

Av 
change of v for this interval is ^- It is the average linear accelerch 

tion^ for this interval. The acceleration at the time t is defined as 
the limit of the average acceleration as At approaches zero. It is 

then ^' The acceleration is the time rate of change of velocity. In 

the case of a falling body it is known experimentally that for 
bodies falUng from heights that are not too great, the velocity 
changes uniformly, due to the action of the force of gravity, i.e., 

• We suppose here that the body is moving in a straight line. If the path is curved, 
it will be seen later that the total acceleration is to be thought of as the resultant of 
two components, one of which produces a change in the direction of the velocity and 
the other a change in the magnitude of the velocity. 



§38] 



THE POWER FUNCTION 



63 



the time rate of change of the velocity is a constant. This con- 
stant is called the acceleration due to gravity and is usually 
denoted by g. In F.P.S. (foot-pound-second) units it is equal 
to 32.2 feet per second per second. That the unit of acceleration 
is 1 foot per second per second is explained by the fact that accelera- 




tion is the change per Second of a velocity of a certain number of 
feet per second. 

The differential equation of motion of the falling body can be 
written 

= 9- (1) 



dt 



From which on integrating, 



V = gt + C. 



If it is given that the body 
starts falling from rest, we have 
as the condition for determin- 
ing C, that V = when t = 0. 
Equation (2) shows that C must 
be equal to zero. Then, 



V = gt. 



(3) 



(2) 




Time U) 

Fig. 27. 



The graph, Fig. 27, o( v = gt is a, straight line whose slope is g. 

If the body had had an initial speed of Vq feet per second, i.e., 

if it had been projected downward instead of being dropped, the 

constant C would have been determined from the condition that 



54 CALCTJLUS [§38 

V = Vo when t = 0. It follows from (2) that C = vo, and the 
equation for v would have been 

V = gt + Vq. (4) 

The graph of this function is shown in Fig. 26. It is again a 
straight line but it cuts the F-axis at the point (0, Vo). 

The foregoing discussion evidently applies equally well to any 
uniformly accelerated motion, i.e., to any motion where the rate of 
change of the velocity is constant. In all such cases the graph of 
r as a function of the time is a straight line. 

Since v = -n' equation (4) gives 

ds 

di = 9i + f 0. 

Integrating, 

s = yt^ + vot + Ci. (5) 

If t is measured from the instant the body begins to move and s 
from the position of the body at that instant, s = when t = 0. 
From this condition d = 0. Then the distance of the body from 
its initial position is given by 

s = igf^ + Vot. 

If a body is thrown vertically upward, it is convenient to count 
distances measured upward, and upward velocities, as positive. 
Then, since the acceleration due to gravity diminishes v, equation 
(1) becomes 

di = -^- (^ ^ 
The formulas (4) and (5) then become 

v^ -gt + vo (4') 

s = - hgt' + Vot. (5') 

If a body falls from rest it is easy to express the speed as a 

function of the distance traversed. In this case, 2^0 = 0. Then 

(4) and (5) become, 

V = gt (4") 



i 



§38] THE POWER FUNCTION 55 

Elimination of t between these equations gives: 

V = VWs- (6) 

Exercises 

1. If a body falls from rest how far will it fall in 10 seconds? 

2. If a body is thrown vertically downward with a velocity of 10 feet 
per second, how far will it have moved by the end of 10 seconds? 
What will its velocity be? 

3. If a body is thrown vertically upward with a velocity of 64.4 
feet per second, what will the velocity be at the end of 10 seconds? 
What will be the position of the body? How far will it have moved? 

4. Find the laws of motion if the acceleration is equal to 2t and if 
(1°) s = and v = when t = 0; (2°) s = 3 and v = --2 when t = 0. 

6. If the acceleration is proportional to the time and ii v = vo and 
8 = So when i = 0, show that 

8 = -^ + Vol + So. 



CHAPTER IV 
DIFFERENTIATION OF ALGEBRAIC FUNCTIONS 

39. The Derivative of the Product of a Constant and a Variable. 

Let 

y = cu, 

where c is a constant and u and y are functions of x. Let Am and 
Ay be the increments of u and y, respectively, corresponding to 
the increment Ax. Then 



2/ + A?/ = 


c{u + Am) 


A?/ = 


cAm 


Ay 

Ax 


Am 
Ax 


dy 

Ix ~ 


du 

c v-» 

dx 


d(cu) 
dx ~ 


du 

C J* 

dx 



or 



The derivative of the product of a constant and a function is equal 
to the constant times the derivative of the function. 
Illustrations. 

d{3x^) d(x2) 

1. — J = O —j — ■• = DX. 

ax dx 



,m__m,_,^j^._,,^_,. 



3. 



d[-f(x 2-5)2] 

dx 





,d{x^ 


^-5)^ 


-i d{x^ - 
dx 




= 


3 
2 


dx 

2-5)- 

X 


A) 




3 

(X2- 

50 


-5)2 





§40] ALGEBRAIC FUNCTIONS 57 

40. The Derivative of the Product of Two Functions. Let 

y = uv, 

where u and v are functions of x. 

y -\-Ay = (u + Au)(v + Aw) 
y + Ay = uv -\- nAv + vAu + AuAv 
Ay = uAv + vAu + AuAv 

Ay 4!i_i_ 4!f_i_A — 
Ax ~ Ax Ax Ax 

Since Aw approaches zero as Ax approaches zero, 

dy _ dv du 

dx dx dx 
or 

d(uv) dv , du ... 

^^dr = ^dx + ^dx' ^^^ 

The derivative of the product of two functions is equal to the first 
times the derivative of the second plus the second times the derivative of 
the first. 

Illustraiions. 

- ^J^i^ = (. + „^J^ + (. + 3,^i^t^ 
= (i + 2) + (I + 3) = 2i + 5. 

= (x2 + 3x) + (x - 2)(2x + 3) 
= 3x2 _|_ 2x - 6. 
3. If x^ + xy^ + y = 10, 

dx ^y ^dx 
2x + 3xy^f^ + y^ + f^ = 0, 

whence 

dy ^ _ 2x + y» ^ 
dx 3xy» + l' 



58 CALCULUS [§41 

Exercises 
Differentiate the following: 



1. (x + l)ix - 1). 9. 2/ = (x -h l)VJi^^. 

2. (x^ + 2x)(x - 3). 10. y = (2 - x)\/^c^~:^. 

3. a;(x2 + 2x - 6). 11. y = {2x ■{■ 3)\/4 - x\ 

4. (x - l)='(x^ + 1). 12. ?/ = x\/n^. 

5. (x' + 2x - 3)(x + 1)2. 13. xV + 3x - 7?/ = 15 

6. xVx - 1. 14. a;2?/ - Sxy^ = lo. 

7. (x - 1)V^. 15. yy/^ + xv^ = 3. 

8. x(x - l)i 16. xy - xhj = 0. 

41. The Derivative of the Quotient of Two Functions. Let 

u 

y = -y 

where u and v are functions of x. Then 

yv = u. 
Differentiating by the rule of §40, 

dv dy _ du 

^ dx dx ~ dz 

du dv 
dy _ dx ^dx 
dx V 



Replacing y by its value, -> 



(v) 



du dv 
Vx ~ ^dx (1) 



dx 



The derivative of the quotient of two functions is equal to the 
denominator times the derivative of the numerator minus the numera- 
tor times the derivative of the denominator, all divided by the square 
of the denominator. 



§42] ALGEBRAIC FUNCTIONS 59 

Illustration. 

47^72-] _ (^ - 2) -^^ (x^ + 1) -^^ - 

dx (a; -2)2 

_ ix-2)i2x) - (x^+1) 
(a: -2)2 
x^ — Ax — 1 



{x - 2)2 

Exercises 

Differentiate the following: 
1. ^- 4. ^^-r-'. 7. 



a;-l 1+a- x - Vx^ - 1 



V'x - 1 


1 -\-x 


x^ - 1 


Vx 


x 



x2 -3 a:2 - 1 g2 4,4 

X — 2 V X •'' — 2 

„ X + x'^ „ a; „a;3 + 8 
3. ~~ 6. T-- 9. hi' 

1 - a; (1 - x)^ ^ ~ ^ 

42. The Derivative of u", n Negative. In Chapter III the 
formula 

dw" , du 

dx dx 

was proved for n positive and commensurable. The formula 
was assumed for negative exponents. We are now in a position 
to give a proof of the formula for this case. Let 





y = u-% 






where s 


is a positive commensurable constant. 


Then 




1 






or 


yw = 1. 






Differentiate by the formula of §40, 








du dy 
ysu'-^^^+u-^^ = 


0, 






dy sy du 

= ■ J- = — su- 

dx u dx 


8-1 ^'^ 

dx 





(1) 



60 CALCULUS [§43 

This completes the proof that 

du° , du 

dx dx 

if n is a commensurable constant.^ 
We see from equation (1) that 



'(J) 



nc du 



dx u°+i dx 

Illustration. 

d S 6 dix^ - 1) 12a; 



(2) 



dx (x2 - 1)2 (a;2 - 1)» dx {x^ - lY 



2. * 





Exercises 




llo 
4. 


wmg: 

3 

X - 1 


7 6 . 

'• x^ + 1 


6. 


5 

(X + ly 


(x^ + D* 


6 


3 


0. ' . 




Vx - 1 


(1-x^y 



1 -X 

Vx 

43. Maximum and Minimum Values of a Fvmction. In Chap- 
ter I it was shown that the derivative of a function with respect to 
its argument is equal to the slope of the tangent drawn to the 
curve representing the function. The derivative is positive where 
the function is increasing and negative where the function ia 
decreasing. These facts enable us to determine the maximum 
and minimum values of a function. 

Additional exercises in finding maximum and minimum values 
of a function will be given in this section. 

Illustration. Let 

J/ = 2x3 + 33-2 _ 12a; - 10. 
• ^ = 6x2 + Gx - 12 = 0(3, _j_ 2) (i _ 1). 

> It can be shown that the formula also holds for incommensurable exponents. 



1843] 
If xi 



ALGEBRAIC FUNCTIONS 



61 



If X is less than —2, both factors of the derivative are negative. 
Then for all values of x less than —2, the derivative is positive 
and the function is increasing. If x is greater than —2 and less 
than 1, the first factor of the derivative is positive and the second 
negative. Hence, if — 2 < x < 1, the derivative is negative and 
the function is decreasing. If x is greater than 1 the derivative is 
positive and the function is again increasing. 

The function changes from an increasing to a decreasing func- 
tion when x passes through the value —2, and changes from a 




decreasing to an increasing function when x passes through the 
value 1. Hence the function has a maximum value when x equals 
—2, and a minimum value when x equals 1. These values, 10 and 
— 17, respectively, are obtained by substituting — 2 and 1 for 
X in the function. (See Fig. 28.) The more important results 
of the above discussion are put in tabular form below. 



X 


X + 2 


X - 1 


dy 
dx 


Function 


X < -2 








+ 


Increasing. 


-2 < I < 1 


+ 


- 


- 


Decreasing. 


1 < X 


+ 


+ 


+ 


Increasing. 


X = -2 





— 





Max. value = 10. 


X = 1 


+ 








Min. value = —17. 



62 CALCULUS [§43 

It is to be observed that — 2 and + 1 are the only values of x 
at which the derivative can change sign and that these are the 
values that need to be examined in finding the maximum and 
minimum values of the function. 

Exercises 

Find where each of the following functions is increasing; decreas- 
ing. Find the maximum and minimum values if there are any. 
Sketch the curve representing each function. 

1. y ^ xK 6. 2/ = (x + 2)(x -3). 

2. y = x\ 1. y = 2a;' - 9x^ + 12x - 10 

3. t/ = -2x\ 8. y = x' - 3x + 7. 

4. y = x' + 3x - 2. 9. 2/ = x' + x2 - X - 1, 

5. 2/ = 3x' - 2x« — 6. 10. y = —^ 

11. A sheet of tin 24 inches square has equal squares cut from 
each corner. The rectangular projections are then turned up to form 
a tray with square base and rectangular vertical sides. Find the side 
of the square that must be cut out from each comer in order that the 
tray may have the greatest possible volume. 

Hint. Show that the function representing the volume of this tray 
is 4x(12 — x)'', where x is the side of the square cut out. 

12. In a triangle whose sides are 10, 6, and 8 feet is inscribed a rec- 
tangle the base of which lies in the longest side of the triangle. Ex- 
press the area of the rectangle as a function of its altitude. Find 
the dimensions of the rectangle of maximum area. 

13. A ship A is 50 miles directly north of another ship £ at a 
certain instant. Ship B sails due east at the rate of 5 miles per hour, 
and ship A sails due south at the rate of 10 miles per hour. Show 
that the distance between the ships is expressed by the function 

V 125/* — lOOOi + 2500, where t denotes the number of hours since the 
ships were in the position stated in the first sentence. At what 
time are the ships nearest together? At what rate are they sepa- 
rating or approaching when f = 3? When < = 5? When < = 6? 

14. The stiffness of a rectangular beam varies as its breadth and as 
the cube of its depth. Find the dimensions of the stiffest beam which 
can be cut from a circular log 12 inches in diameter. 



§431 ALGEBRAIC FUNCTIONS 63 

Miscellaneous Exercises 

Differentiate tlie following twenty-five functions with respect to x: 



l.i. 




9. (3 - x)^ 


18. 


X2(l - X). 




X 




10. (2 - x^y. 


19. 


x(i - x^y. 






11. (3 - x=')-2. 


20. 


(1 - X3)2(X 


-2)2. 


4. (X - sy. 




12. (2 - x*)-". 

13. (x + l)(x - 2). 

14. Vi. 


21. 
22. 


X - 1. 
X2 + 1 
1 -X 

1 +X2 




5. (x - 2)^ 




15. x\ 


23. 


(x - l)i 




6. (x2 - 1)2. 


x-6) 


16. ]. 


24. 


X 




7. (x' - 2x2 + 


Va^ - x2 




8. (x - 2) -3. 




17. x(x - 1)2. 


26. 


XVI - X2. 




Integrate the 


following twenty expressions : 








26. ^ = x3. 

dx 




37. '^ .-? 

ax X 


i' 







27. J = x3+x2. . 38. J = ^^^:^ 

28. -^-^ = 0.3 + x2 + X + 1. 39. ^^ ^ 



^^ - .c -r u. -r -, -r X. ^^ ^^ _ ^^^^ 

^^ dy 1 .^ dy \ — X 

29. -7^ = x2. 40. - - 



dx ~ dx (2x - x2)2 

30. ^=-4- 41.^^- ^ 



dx ^x ' ^^ Vx + 1 

31. :7: == a;^ + x^. 42. ■;t- = 



dx ' ' dx "v/x^-T 

a.. I = (. - 1,^. 43. g - ^^. 

44. —- = x"i 
dx 

46. ^ = X - 1. 

dx 

46. Find ~\i x^ - y"^ = a\ 

^-Ftodgitfa + l'-i. 

48. A ladder 20 feet long leans against the vertical wall of a 



33. 


dy 

dx ~ 


x(l - 


X2)2. 


34. 


dy 

dx ~ 


X2(x3 


- 2)5 


36. 


dy 
dx ~ 


1 




(1- 


xy 


36. 


dy 

dx ~ 


1 

X2' 





64 



CALCULUS 



:§43 



building. If the lower end of the ladder is drawn out along the 
horizontal ground at the rate of 2 feet per second, at what rate is 
its upper end moving down when the lower end is 10 feet from 
the wall? 

Hint. Let AC, Fig. 29, be the wall and let CB be the ladder. 
I-et AB =x and AC = y. Then 



and 



V 

dy 
dt 



= \/400 - 



But, since 



dx 
dt 

dy 
dt 



- X dx 
\/400 - x^ dt' 



■-2X 

^400 - x^ 




The negative sign of the deriva- 
tive indicates that the upper end pjQ 29. 
of the ladder is moving down. 

49. Answer the question of Exercise 48, ifx=0;x=2;a; = 15; 
a; = 20. 

50. With the statement of Exercise 48, find the rate at which the 
area of the triangle ABC, Fig. 29, is increasing when the lower end 
of the ladder is 5 feet from the wall. 

51. With the statement of Exercise 48, find the position of the ladder 
when the area of the triangle ABC, Fig. 29, is a maximum. 




Fig. 30. 

52. A ball is dropped from a balloon at a height of 1000 feet. Ex- 
press the velocity of its shadow along the horizontal ground as a 
function of the time, if the altitude of the sun is 20°. 

Hint. Let x. Fig. 30, be the distance of the falling body above 
the earth. Let y be the distance of the shadow from a point on the 
earth directly under the falling body. 



§43] ALGEBRAIC FUNCTIONS 65 

53. With the statement of Exercise 52, find the velocity of the 
shadow when the ball leaves the balloon; when it is half way to the 
earth; when it reaches the earth. 

64. A man standing on a dock is drawing in a rope attached to a boat 
at the rate of 12 feet per minute. If the point of attachment of the 
rope is 15 feet below the man's hands, how fast is the boat moving 
when 13 feet from the dock? 

55. The paths of two ships A and B, sailing due north and east, 
respectively, cross at the point C. A is sailing at the rate of 8 miles 
per hour, and B at the rate of 12 miles per hour. If A passed through 
C 2 hours before B, at what rate are the two ships approaching or 
separating 1 hour after B passed through C? 3 hours after B passed 
through C? When are the two ships nearest together? 

56. Two bodies are moving, one on the axis of x, the other on the 
axis of y, and their distances from the origin are given by 

X = 31^ - 31 + 1, 
y = 6t - 12, 

the units of distance and time being feet and minutes, respectively- 
At what rate are the bodies approaching or separating when 1 = 2? 
When t = 5? When are they nearest together? 

57. A ship is anchored in 35 feet of water and the anchor cable 
passes over a sheave in the bow 15 feet above the water. The cable 
is hauled in at the rate of 30 feet a minute. How fast is the ship 
moving when there are 80 feet of cable out? 

58. A gas in a cylindrical vessel is being compressed by means of a 
piston in accordance with Boyle's law, pv = C. If the piston is 
moving at a certain instant so that the volume is decreasing at the 
rate of 1 cubic foot per second, at what rate is the pressure changing if 
at this instant the pressure is 5000 pounds per square foot and the 
volume is 10 cubic feet? 

59. Water is flowing from an orifice in the side of a cylindrical tan k 
whose cross section is 100 square feet. The velocity of the water in 
the jet is equal to \/2gh, where h is the height of the surface of the 
water above the orifice. If the cross section of the jet is 0.01 square 
foot, how long will it take for the water to fall from a height of 100 
feet to a height of 81 feet above the orifice? 

60. At a certain instant the pressure in a vessel containing air is 
3000 pounds per square foot; the volume is 10 cubic feet, and it is 
increasing in accordance with the adiabatic law, piP-* = c, at the rate 
of 2 cubic feet per second. At what rate is the pressure changing ? 

6 



66 CALCULUS [§44 

61. Water flows from a circular cylindrical vessel whosa radius is 
2 feet into one in the shape of an inverted circular cone whose vertical 
angle is 60°. (a) If the level of the water in the cylinder is falling 
uniformly at the rate of 0.5 foot a minute, at what rate is the water 
flowing? (6) At this rate of flow, at what rate will the level of the 
water in the cone be rising when the depth is 4 inches? When it is 
20 inches? 

62. A toboggan slide on a hillside has a uniform inclination to the 
horizon of 30°. A man is standing 300 feet from the top of the slide 
on a line at right angles to the slide. How fast is the toboggan moving 
away from the man 3 seconds after leaving the top? 10 seconds after 
leaving the top? (Use formula for speed of a body sliding down an 
inclined plane. Neglect friction.) 

If the man is approaching the top of the slide at the rate of 10 feet 
a second, answer the same questions, it being supposed that the man is 
300 feet away from the top of the slide when the toboggan starts. 

44. Derivative of a Function of a Function. If y = <j)(u) and 
u = fix), ?/ is a function of x. The derivative of y with respect 
to X can be found without eliminating u. For any set of corre- 
sponding increments, Ax, Ay, and Au, 

Ay _ Ay Au 
Ax Au Ax 
Hence 

lim A^ ^ lim 4^ lim Aw^ 

Since Am approaches zero as Ax approaches zero, 

dx du dx ^ ' 

This is the formula for the derivative of a function of a function. 
Illustration. Let 

y = u^ -\- 5 
and 

u = 3x^ + 7x + 10. 

du 
and 



§46J ALGEBRAIC FUNCTIONS 67 

Then 

^ = 3m2(6x + 7) 

= 3(3a;2 + 7x + 10)^ (6x + 7). 

dv 
45. Inverse Functions. If x = 't>iv),li^ can be found by the rule 

^ = A, 

dx dx 

dy 
which is easily proved. 

dy ^ Mm ^ ^ Mm i_ ^ ±, 
dx '^^-0 Ax ^x^^Ax dx 
Ay dy 
Illustration. U x = 5y^ -\- 7y^ + 3, 

^^ = 302/« + 147/, 



and 



dy 

dy 1 

dx 2y{lby* + 7) 



Exercises 



1. Find-TT- in terms of X if: 
ax 



(a) y = Vu^+ 7 and m = 3a; + 10. 
(6) 7/ = 2«3 + 5m and i* = x" - 2x. 

(c) y = — and w = x^ — 2. 



2. Find ^ if 
dx 




W ^ - (7/2+2)2 



3. Find ^ if: 
dx 



(a) y* +x* - 7xy = 15. (6) Sxy^ + &x'y + 4x2 = 15. 
46. Parametric Equations. If the equation of a curve is given 
in parametric form, x = f{t), y = (f>{t), it is important to be able 
to find the derivative of y with respect to x without eliminating « 



68 



CALCULUS 



[§47 



between the given equations. A rule for doing this can be derived 
by the method used in §§13 and 17. 

If I is given an increment At, x and y take on the increments 
Ax and Ay, respectively. Then 

Ay 
Ay At 



and 



or 



Ax 


Ax 




At 




Um At/ 


lim ^y 


At=0 ^t 


M=0 ^x 


lim Ax 




At = ^{ 




dy 


dy 
dx 


dx 




dt 



Exercises 

1. Find the slope at (6,1) of the curve whose parametric equations 
arc 

X = t^ +t, 

2/ = < - 1. 
Find -T- for each of the following: 

2. x = t", 

2/ = i^ + 1. 

3. x = u^ +3, 
1 

47. Lengths of Tangent, Normal, Subtangent, and Subnormal. 

In Fig. 31, PT is the tangent and FN is the normal at P. The 
lengths of the lines PT, PN, TD, and DN are called the tangent, 
the normal, the subtangent, and the subnormal, respectively, for 
the point P. Show that the lengths of these lines are; 

y^ 

¥ (1) 

dx 




y 



TD = 



§47] ALGEBRAIC FUNCTIONS 69 

DN = y'£. (2) 



"^-UHir- 



(3) 



dx 



=4- (I) 



PN = yjl -h (^] • (4) 



Exercises 

1. Obtain the length of the tangent, normal, subtangent, and sub- 
normal for the point (1, 2) on the curve y* = 4x. Show that for points 
on this curve the subnormal is of constant length. 
• 2. Write the equation of the tangent to y^ = 4x at the pomt (1,2). 
Write the equation of the normal at the same point. It is to be noted 
that in this exercise the equations of the tangent and normal lines 
are to be found, and not the lengths of the tangent and normal as 
in the preceding exercise. 

3. Write the equation of the tangent to 

I* «* 

— +— = 1 

25 ^9 ^ 

at the point (3, 2.4). Use implicit differentiation. 

4. Find the equation of the curve whose subnormal is of constant 
length 4 and which passes through the point (1, 3). 

2 
6. Find the length of the tangent to j/ = - at the point where 

y = 1. 

6. Find the length of the normal to the curve y ■ 
point where x = 3. 

7. Find the equation of the curve passing through the point (1, 3) 
and having a subtangent equal to the square of the ordinate. 



CHAPTER V 

SECOND DERIVATIVE. POINT OF INFLECTION 

48. Second Derivative, Concavity. Since the first derivative 
of a function of x is itself a function of x, we can take the deriva- 
tive of the first derivative. The derivative of the first derivative 
is called the second derivotive. In the case of a function y of x, it 

di^) 
is denoted by the symbol — -r- — , or -r- vj') , or more commonly by 

dh/ dhi 

-j-j* -r-^ is read "the second derivative of y with respect to x." Ifr 

d^ 
must here again be remembered that -r-^ is not a fraction with a 

numerator and a denominator, but is only a symbol representing 
the derivative of the first derivative. 

If y = f{x), the first derivative of y with respect to x is some- 
times written y' and very commonly fix). Similarly the second 
derivative is indicated by/"(a;). 

The derivative of the second derivative is called the third deriva- 

d^v 
tive. It is designated by -r-^» or ii y = f(x), by f"'{x). The nth 

derivative is designated by-i-^' or by/^"^^(x). 

Between the points A and C, Fig. 16, where the curve is con- 
cave downward, the slope of the tangent decreases from large 
positive values near A to negative values near C. This means 
that the tangent revolves in a clockwise direction as the point of 
tangency moves along the curve from A toward C. Clearly this 
will always happen for any portion of a curve that is concave 
downward. (See Fig. 32, a, h, and c.) The slope decreases as the 
point of tangency moves to the right. 

On the other hand, if a portion of a curve is concave upward, 
the slope of the tangent increases as the point of tangency moves 
to the right. Thus in Fig. 16 the slope of the tangent is negative 
at C and increases steadily to positive values at E. The same 

70 



§49] SECOND DERIVATIVE 71 

thing is evidently true for any portion of a curve that is concave 
upward. In this case the tangent line revolves in a counter- 
clockwise direction. 

Since the first derivative of a function is equal to the slope of 
the tangent to the curve representing the function, what has just 
been said can be stated concisely as follows: 

If an arc of curve is concave upward the first derivative is an 
increasing function, while if the curve is concave downward, the 
first derivative is a decreasing function. 

If the second derivative of a function is positive between cer- 
tain values of the independent variable x, the first derivative is an 
increasing function, the tangent line revolves in a counter- clock- 





FiG. 32. 

"wise direction, and consequently the curve representing the 
function is concave upward between the values of z in question. 
If the second derivative is negative, the first derivative is a de- 
creasing function and the curve is concave downward. Thus in 
Fig. 16 the second derivative is negative between A and C, and 
between E and G. It is positive between C and E, and between 
G and /. 

49. Points of Inflection. Points at which a curve ceases to be 
concave downward and becomes concave upward, or vice versa, 
are called 'points of inflection. 

At such points the second derivative changes sign. C, E, and 
G, Fig. 16, are points of inflection. At C, for instance, the second 
derivative changes from negative values to positive values. 

Illustration 1. Study the curve y = |x' by means of its 
derivatives. 

Differentiating, 

dx~ ^^* 

d^y _ 



72 



CALCULUS 



[§49 



When a; < 0, -^ < 0, -7- = ^x'^ is a decreasing function, and 

the curve y = |x' is concave downward. When x > 0, 

d^y dy . . 

-Y~2 ^ ^> TT ^^ ^^ increasing function, and the curve y = Ix^ is 

concave upward. 




Fig. 33. 



d'^y 

At the point where x = 0, -7^2 changes sign from negative to 
positive, and the curve changes from being concave downward to 



SECOND DERIVATIVE 



73 



being concave upward 
inflection. 
dy 



Hence the point (0, 0) is a point of 



Since ^- is positive except when x = 0, ?/ 



iX* IS an in- 



creasing function excepting when x = 0. When x = the curve 
has a horizontal tangent. 




Fio. 34. 



In Fig. 33 the graphs of the functions y= \x^ and of its first 
and second derivatives are drawn on the same axes. Trace out 
in this figure all the properties mentioned in the discussion. 



Illustration 2. 


Let 










y = 


Ix' 


— x^ -\- Ix 


+ 2. 


Differentiating, 


J 










dy _ 
dx 


Ix- 


-2x + l 






d'y 
.dx^ 


h(x 
X — 


- l){x - 
2. 


3). 



74 



CALCXJLUS 



[§49 



At X = 2, -T^ changes sign from minus to plus. Hence the curve 

is concave downward to the left, and concave upward to the right 
of the line x = 2. The point on the curve whose abscissa is 2 is 
then a point of inflection. The value of the function corresponding 
to X = 1 is a maximum value, and the value of the function 
corresponding to x = 3 is a minimum value. See Fig. 34 for a 
sketch of the function and its first and second derivatives. Trace 
out in the figure what has been given in the discussion. 

The more important properties of the function are put in tabular 
form below. 



X 


dx^ 


dy 
dx 


Curve 


X < 2 
X > 2 
X = 2 


+ 



Decreasing 
Increasing 


Concave downward. 

Concave upward. 

Point of inflection {y = 2^), 


X < 1 
1 < X < 3 
X > 3 
X = 1 
X = 3 




+ 

+ 




Increasing, 

Decreasing. 

Increasing, 

Maximum point (v = 2f). 

Minimum point {y = 2), 



Exercises 

Find the maximum and minimum points and points of inflection of 
the following curves. Sketch the curves, 

1. y = x' — 3x^ 

2. y = x' 4- 3x2, 

3. 2/ = 2x3 + 3x2 4. ga; _|_ 1, 

4. 2/ = 3x* - 4x3 - 1. 
6. y = x'. 

6. t/ = 2x< - 4x3 - 9x2 _^ 27x + 2, 
1. y = 6x^ - 4x3 _|_ I 



CHAPTER VI 



APPLICATIONS 



60. Area under a Curve: Rectangular Coordinates. An im- 
portant application of the anti-derivative is that of finding the 
area under a plane curve. 

Let APQB, Fig. 35, be a continuous curve between the ordinates 
X = a and x = b. Further, between these limits, let the curve 
lie entirely above the X-axis. Our problem is to find the area, 
A, bounded by the curve, the X-axis, and the ordinates x = a 
and X = h. 




The area can be thought of as generated by a moving ordinate 
starting from x = a and moving to the right to a position DP where 
the abscissa is x. This ordinate sweeps out the variable area u, 
which becomes the desired area A when x = b. On moving from 
the position DP to the position EQ where the abscissa is a; + Ax, 
the ordinate to the curve takes on an increment Ay and the area u 
an increment Aw. By taking Ax small enough the curve is either 
ascending or descending at all points between P and Q. It 
follows at once from the figure that 

yAx < Au < (y -{-Ay)Ax, (1) 



or 



Au , . 

y <x:;.<y + ^y- 



Ax 



75 



76 CALCUI.US [§50 

(If the curve descends between P and Q the signs of inequahty in 

(1) are reversed. The argument which follows will not be affected.) 

As Ax approaches zero, Ay approaches zero and y -\- Ay 

approaches y. Hence v-' which lies between y and y + Ay, 

approaches y. Thus 

lim Am ^ 

Aa:=0 J>^x ^' 



or 






If the equation of the curve is y = f(x), 

g = m. (3) 

Let F{x) be a function whose derivative is /(x) . Then 

u = Fix) + C. 

C is determined by the condition that u = when x = a. Then 

C = - F(a) 
and 

u = F{x) - F(a), (4) 

an expression for the variable area measured from the ordinate 
X = a to the variable ordinate whose abscissa is x. A, the area 
sought, is obtained by putting x = 6 in equation (4). 

A = F(b) - F(a) (5) 

Illustration. Find the area A bounded hy y = x', the X-axis, 
and the ordinates x = 2 and x = 4. 













du 
dx 


= x\ 














u 


= \X' 


' + c. 


When X 


= 2, 


u 


= 0, 


andC 


= - 


- A 
3* 


Then 


and 










u = 


\x' 


_ 8 
i) 



§51j APPLICATIONS 77 

Exercises 

1. Find the area bounded by the X-axis, the lines x = 1 and x = 2, 
and 

(a) y = mx. 

(b) y = x^. 

(c) y = 2x2 + 3x + 1. 

2. Find the area between the curves y = x^ and x = j/'; 
. y* = a{a — x) and y = a — x; y^ = ^x and y = 2x; y^ = x* and ?/ = x. 

3. Find the area bounded hy y = ■%/ x + 1, the Z-axis, and the 
ordinates x = and x = 2. 

51. Work Done by a Variable Force. In this section there is 
given a method of finding the work done by a variable force 
whose line of action remains unchanged. 

Illustrations of such variable forces are: 

1. The force of attraction between two masses, m and ilf, is 

given by the Newtonian law 

ItMrn 
fis) = -^' 

where s is the distance between the masses and A; is a factor of 
proportionality. Note that the equation is of the form 

f(s) - "-,■ 

2. The force exerted by the enclosed steam on the piston of a 
steam engine is, after cut-off, a function of the distance of the 
piston from one end of the cylinder. 

3. The force necessary to stretch a bar is a function of the 
elongation of the bar. 

Let AB, Fig. 36, represent a bar of length I, held fast at the 
left end, A. A force / is applied at its right end and the bar 
is stretched. It is shown experimentally that up to a certain limit 
the elongation, s, is proportional to the force applied (Hooke's 
Law), i.e., 

f = ks, 

where A; is a constant depending upon the length of the bar, 
its cross section, and the material of the bar. 



78 



CALCULUS 



[§51 



The work done by a constant force in producing a certain 
displacement of its point of application in its line of action is 
defined as the product of the force by the displacement. In 
the problem which we are considering the force varies with the 
displacement. The work cannot be found by multiplying the 
displacement by the force. Instead it will be found by integrat- 
ing an expression for the derivative of the work with respect 
to the displacement. 

Let w denote the work done in producing the displacement from 
s = a to a variable position s = s. Let Aw denote the work 
done in producing the additional displacement As. Let / denote 
the force acting at s, and / + A/ the force acting at s + As. A/ 




I I 



FiG. 36. 



may be positive or negative according as the force increases or 
decreases with distance. For definiteness suppose A/ positive. 

In producing the displacement As the force varies from / to 
/ + A/, and hence the work Aw; lies between /As and (/ + A/) As, 
which represent the work which would have been done had the 
forces / and / + A/, respectively, acted through the distance As. 

Hence 

/As < Aw; < (/ + A/) As, 



or 



f<fs<f+^f- 



As As approaches zero, A/ approaches zero, and we obtain, 

dw 



ds 



= f. 



(1) 



§51] APPLICATIONS 79 

Integration gives 

w = F{s) + C, (2) 

where F{s) is a function whose derivative is /. 

When s = a, w = Q, (2) gives C = — F{a) and, 

w =F{s) -Fia). (3) 

This represents the work done in the displacement from s = a 
to s = s. The work, W, done in the displacement from s = a 
to s = & is obtained by substituting 6 for s in (3). 

W = F(b) - F(a). (4) 

Illustration 1. Find the work done in stretching a spring from a 
length of 20 inches to a length of 22 inches, if the length of the 
spring is 18 inches when no force is applied and if a force of 30 
pounds is necessary to stretch it from a length of 18 inches to a 
length of 19 inches. 

Denote the elongation of the spring by s. In accordance with 
Hooke's Law, 

/ = ks. 

Since s = 1 when / = 30, A; = 30 and/ = 308 
Substituting in equation (1), 

-r = 30s. 
as 

w = 15s2 + C. 

The problem is to find the work done in changing the elonga- 
tion from s = 2 to s = 4. When s = 2,w = 0. Hence C = — 60, 
and 

w = 15s2 - 60. 

The required work, W, is found by giving to s the value 4. 

TF = 240 - 60 = 180. 

Thus the work done is 180 inch-pounds, or 15 foot p junds. 

Illustration 2. Two masses M and m are supposed concen- 
trated at the points A and B, respectively. Find the work done 



80 



CALCULUS 



[§51 



against the force of attraction in moving the mass m along the line 
AB from a distance a to a distance b from the mass M, the latter 
mass being fixed. 

If / is the force of attraction between the two masses, 

_ kmM 

f - g2 

Then 

dw _ kmM 
Is ~~ s2 
From which 

kmM 



w = — 



+ C. 



-S +AS- 



FiG. 37. 



When s = a, w = 0. Hence 




kmM 




and 




w = kmM — 
\_a 


1- 


To find the required work, PF, let s = 6, 




W = kmM - - 
[a 


1- 
b_ 



Illustration 3. Gas is enclosed in a cylinder, one end of which is 
closed by a movable piston. Find the work done by the gas in 
expanding in accordance with the law pv^-* = K, from a volume 
of 3 cubic feet at a pressure of 15,000 pounds per square foot to 
a volume of 4 cubic feet. 

Let A be the area of the cross section of the cylinder. Then 
pA is the force on the piston, Fig. 37, and 

dw 
ds=P^' 



§51] 






APPLICATIONS 


or 












dw 


AK AK K I 






ds 


~ v^* ~ (AsY* ~ A'^U^*' 


Integration i 


gives 


1 K 1 

^~ 0.4 ^""s"'*'^^ 








- ^ ^ +C 


When V = 3 


,w 


= 0. 


Hence 

1 K 
0.4 3°"' 


and 






K r 1 1 -] 
^ 0.4 [s^* v°*y 

0.4 [3°" 40''_ 



81 



When t; = 3, p = 15,000. Hence K = (15,000)(3i"), and 

W = 150,000[0.75 - (0.75) i«l 
= 12,230. 

Exercises 

1. A spring is 12 inches long and a force of 120 pounds is necessary 
to stretch it from its original length, 12 inches, to a length of 14 inches. 
Find the work done in stretching the spring from a length of 13 inches 
to a length of 15 inches. 

2. In the case of a bar under tension, Fig. 36, the relation between 
the stretching force, /, the original length of the bar, I, and the elonga- 
tion of the bar, s, is given by 

^ EAs 

^ = -r' 

where E is the modulus of elasticity of the material of the bar and 
A is the area of the cross section of the bar. Find the work done in 
stretching a round iron rod ^ inch in diameter and 4^ feet long to a 
length of 54.5 inches, given that E = 3-10^ pounds per square inch. 

3. A spherical conductor, A, is charged with positive electricity and 
a second spherical conductor, B, with negative electricity. The force 
of attraction between them varies inversely as the square of the 

6 



82 



CALCULUS 



[§52 



distance between their centers. If the force is 10 dynes when the 
centers are 100 centimeters apart, find the work done by the force of 
attraction in changing the distance between the centers from 140 
centimeters to 120 centimeters. 

4. Find the work done by a gas in expanding in accordance with 
the law pv^ = C from a volume of 5 cubic feet to one of 6 cubic 
feet, if p = 70 pounds per square inch when v = 5 cubic feet. 

5. Find the work done in compressing a spring 6 inches long to a 
length of 5§ inches if a force of 2000 pounds is necessary to compress 
it to a length of 5 inches. 

6. The work done by a variable force can be represented graphically 
as the area under a curve whose ordinates represent the force. Con- 
struct the figures and prove this fact for Illustrations 1, 2, and 3. 

62. Parabolic Cable. Suppose a cable, AOB, Fig. 38, a, is 
loaded uniformly and continuously along the horizontal, i.e., so 
that any segment of the cable sustains a weight proportional to 




the projection of the segment upon a horizontal line. Let k be the 
weight carried by a portion of the cable whose horizontal pro- 
jection is one unit of length. 

Choose 0, the lowest point of the cable, as origin and a hori- 
zontal line through as axis of x. Let P be any point on the cable. 
Suppose the portion OP of the cable cut free, Fig. 38, b. To keep 
this portion in equilibrium a horizontal force H and an inclined 
force T must be introduced at the points and P, respectively. 
The force H must be equal in magnitude to the tension in the cable 
at 0, and it must act in the direction of the tangent line at that 
point. Similarly, the force T must be equal to the tension in 
the cable at the point P and act in the direction of the tangent 
line. The force T can be resolved into its vertical and horizontal 



§53] APPLICATIONS 83 

components V and H', respectively. Now H and H' are the only 
horizontal components of the forces acting on OP and, since OP is 
in equilibrium, they must balance each other. Therefore, 

H = H'. (1) 

Hence the horizontal component of the tension in the cable is 

independent of the point P, i.e., it is a constant. 

In like manner the only vertical components of the forces 

acting on OP are the weight kx supported by OP, acting downward, 

and V, the vertical component of T. They must balance one 

another. Hence 

V = kx. (2) 

V 
The slope of the tangent Ime to the curve at the point P is „/ 

Then 

dy _ V^ _kx 

dx~ H'~ H' ^'^' 

This is the slope of the curve at any point. On integrating 
we obtain the equation of the curve apart from the arbitrary 
constant C. 

y = % + c. (4) 

C is determined by the condition that y = when x = 0. Then 
C = 0, and (4) becomes 

y = 2^- (5) 

This is the equation of a parabola with its vertex at the origin. 

63. Acceleration.^ In §38 acceleration was defined as the time 
rate of change of velocity, i.e., as the derivative of the velocity 
with respect to the time. But velocity is the derivative of dis- 
tance with respect to time. Hence the acceleration is the second, 
derivative of the distance with respect to the time. If s denotes 

the distance and t the time, the acceleration is expressed by -w-. 
In the case of a freely falling body 

d'^s „. 

• The statements in this section refer to motion in a straight line. 



84 CAIiCULUS [§53 

The relation between s and t can be found from this differential 
equation by integrating twice, as follows: 
The first integration gives 

and the second 

s = igt^ + Cit + C2. (3) 

Two arbitrary constants of integration are introduced. They 
can be determined by two conditions. If 

s = So (4) 

and 

v = -^ = Vo (5) 

when t = 0, (2) gives Ci = Vq, and (3) gives C2 = Sq. Then 

s = hgt' + vot + So. (6) 

This result was found in §38 by essentially the same method, where 

dv d^s 

the symbol n^ was used instead of -n^' 

Exercises 

1. Solve Exercise 5, §38, by the method used above. 

2. Obtain the relation v = \/2gs (see §38) directly from the 
equation 



ds 
Hint. Multiply by 2 -jr • 



dt^ 



= 9- 



ds d^s _ n f^s 
^dt di^ ~^^dt' 

(ds\i 
~T7 I with respect to t and the 

second that of 2gs. We then have 

Determine C by the condition that v = Q when 8 = 0. 



§54] 



APPLICATIONS 



85 



64. The Path of a Projectile. An interesting application of 
integration is to find the equation of the path of a projectile, a 
baseball for instance, thrown with a given velocity at a given 
inclination to the horizontal. 

Let 0, Fig. 39, be the point from which the ball is thrown. 
Take this point as the origin of a system of rectangular coordi- 
nates. Let the ball be thrown so that its direction at the instant 
of leaving the hand makes an angle a with the horizontal, and let 
the initial velocity of the ball be Vq. Then the horizontal com- 
ponent of the initial velocity is Vo cos a, while its vertical 
component is Vo sin a. That is, at the instant the ball is thrown 
its x-coordinate is increasing at the rate of Vo cos a feet per second. 
Similarly the initial rate of change of the ^/-coordinate is vq sin a. 




Fig. 39. 



At the end of t seconds after the ball was thrown it is at the point 
P whose coordinates are x and y. If the resistance of the air is 
neglected there is no force acting on the ball tending to change the 
component of its velocity parallel to the X-axis. Hence the 
x-component of the velocity is at all times the same as at the 
beginning, viz., Vq cos a. The x-component of the velocity is also 

dx 

-rr, viz., the time rate of change of the abscissa of the ball. There- 
fore we can write 

dx . . 

jT = Vo cos a. {1) 

From which on integration 

X = {vo cos a)t + C. (2) 

Time is counted from the instant the ball was thrown. The 



86 CALCULUS [§54 

condition for determining C is then that a; = when f = 0. It 
follows that C = 0, and (2) becomes 

X = {vo cos a)t. (3) 

This equation gives the x-coordinate of the ball at any time t. 

In the vertical direction, the force of gravity acts to change the 
t/-component of the velocity. 
Then 

g= -.. (4) 

The negative sign is used since the force of gravity causes the 
velocity in the direction of the positive F-axis to decrease. In- 
tegration gives 

1= -gl + C. (6) 

dv 
d is determined by the condition that 37 = ^0 sin a when t = 0. 

Then C2 = Vq sin a and (5) becomes 

^y = — gt + v^ sin a. (6) 

Integrating again, 

y = - W + i^o sin ci)t + C3. (7) 

Since y — when ^ = 0, C3 = 0, and (7) becomes 

y = — hgi^ + (^0 sin a)L (8) 

This is the ^/-coordinate of the ball at any time t. Equations (3) 
and (8) are the parametric equations of the path of the ball. 
The elimination of t between these equations gives the rectangular 
equation of the path, 

y = — :r~2 r~ x^ + x tan a. (9) 

" 2^0 cos^ a 

This is the equation of a parabola with its vertex at the point, 

r?;o^ sin 2a Vq'^ sin^ a"! 
L 2g ' ~2^J" 

Tt is to be remembered that in the solution of this problem the 



§54] APPLICATIONS 87 

resistance of the air was neglected. Consequently the results 
obtained can be regarded only as approximations. Experimentally 
it has been shown that the resistance of the air increases with the 
velocity of the moving body. For low velocities the resistance is 
assumed to vary as the first power of the velocity, but for higher 
velocities, such as are attained by rifle balls, the resistance is as- 
sumed to vary with the second power, and the results obtained 
above cannot be considered to be even approximations. 

Exercise 

1. Find the angle of elevation, a, at which the ball must be thrown 
to make the range, OA, Fig. 39, a maximum. 



CHAPTER VII 
INFINITESIMALS, DIFFERENTIALS, DEFINITE INTEGRALS 

55. Infinitesimals. In §23 an infinitesimal was defined as a 
variable which approaches the limit zero. Thus, x^ — 1, as z 
approaches 1, is an infinitesimal. 

It is to be noted that a variable is thought of as an infinitesimal 
only when it is in the state of approaching zero. Thus x^ — 1 is 
an infinitesimal only when x approaches +1 or —1. An in- 
finitesimal has two characteristic properties: (1) It is a variable. 
(2) It approaches the limit zero; i.e., the conditions of the problem 
are such that the numerical value of the variable can be made less 
than any preassigned positive number, however small. 

This meaning of the word infinitesimal in mathematics is entirely 
different from its meaning in everyday speech. When we say 
in ordinary language that a quantity is infinitesimal, we mean 
that it is very small. But it is a constant magnitude and not one 
whose numerical measure can be made less than any preassigned 
positive number, however small. Thus, 0.000001 of a miUigram 
of salt might be spoken of as an infinitesimal quantity of salt, but 
the number 0.000001 is clearly not an infinitesimal in the sense 
of the mathematical definition. On the other hand, if we have a 
solution containing a certain amount of salt per cubic centimeter 
and allow pure water to flow into the vessel containing the solu- 
tion while the solution flows off through an overflow pipe, the 
quantity of salt per cubic centimeter constantly diminishes. The 
amount of salt left in solution after a time t is then an infinitesimal, 
as t becomes infinite. 

Infinitesimals are of fundamental importance in the Calculus. 
The derivative, which we have already used in studying functions, 
is the limit of the ratio of two infinitesimals. Ay and Ax. 

56. ^f^ • Let the arc AB, Fig. 40, subtend an angle a 

88 



INFINITESIMALS 



89 



at the center, O, of a circle of radius r. The angle a is measured 
in radians. Let AT he tangent to the circle at A, and let BC be 
perpendicular to OA. The area of the triangle OCB is less than 
the area of the circular sector OAB, and this in turn is less than 
the area of the triangle OA T. 

UBC){OC) < har^ < ^iAT)r 

BC OC ^ ^AT 
— — - < a < 

r r r 

OC . 

■ — sin a < a < tan a 
r 



OC 



< 



sin a 

As the angle a approaches zero, 

OC 
OC approaches r and — approaches 

1, and further, cos a approaches 1. 
Hence the first and last members 
of the inequalities (2) approach the 
same limit, 1. Then the second 

a 

member, -; — ■» which lies between 
' sin a 

them, must approach the same limit, 1. Therefore 




lim sin a 



lim 

a=0 



1 

a 

sin a 



= I. 



(3) 



57. 



lim tan a lim tana 
sin a 

tan a 



a=0 



lim 



a 



^ lim /sina 1 \ 
«-o \ a cos a/ 

^ Aim sina\ /lim 1 \ 
\a=0 cc / \°'=° cos OC/ 



= 1. 



lim tan a 



lim 

a=0 



= 1. 



(1) 
(2) 



58. 



lira 1 — cos a 



1=0 



90 CALCULUS [§59 

Since 

^ 2 sin'' fT sin 7, 

1 — cos a 2 2 . a 



= = sin ^» 

a a a 2 



lim l-coso! /lim ^^" 2 W IJm . a 
cc^Q a ~ I «=^o a / I «^o ^ 2 

Hence 

lim 1 — cos « _ 

a=^o « ~ "• ^^^ 

In Fig, 40, AB, AC, AT, BT, and BC are infinitesimals as a 
approaches zero. Then, 

from (3), §56, i™ ^ = h 

from (1), §57, i^o ^ = 1, 

from (2), §57, i^o ^^ = 1, 

from (1), §58, i^o J^ = 0. 

59. Order of Infinitesimals. Consider the infinitesimals x^ 
and X as X approaches zero. The ratio of x^ to x is x, which is 
itself an infinitesimal. The infinitesimals x^ and x are repre- 
sented. Fig. 41, by the ordinates MP and MN, to the curves 
y = x^ and y = x. The quotient 

X ~ MN 

is a measure of the relative magnitude of these infinitesimals as 
they approach zero. It shows that MP becomes small so much 
more rapidly than MN that the limit of their quotient is zero. 
On the other hand, the infinitesimals 2x and x behave very 

2x 
differently. Their quotient is — = 2, and the limit of this 



J59] 



INFINITESIMALS 



91 



quotient is 2. In this case the limit of the ratio of the infini- 
tesimals is not zero. (See Fig. 42.) 
Again, 

lim 1 — c os a 

while 



= 0, 



lim sma 

«-0 rv 



= 1. 



These illustrations of the comparison of two infinitesimals lead 
to the following definitions of the order of one infinitesimal with 
respect to another. 

Two infinitesimals, a and /3, are said to be of the same order if the 

ex 

limit of ^ is a finite number not zero. 




Fig. 41. 



Fig. 42. 



7/ the limit of ^ is zero^ a is said to be of higher order than jS. 

Thus, 2x and x are of the same order; x^ is an infinitesimal of 
higher order than x; sin a and a, or CB and AB, Fig. 40, 
are of the same order; tan a and a, or AT a,nd AB, Fig. 40, are of 
the same order; tan a and sin a, or CB and AT, Fig. 40, are of the 
same order; 1 — cos a is of higher order than a, or CA, Fig. 40, 
is of higher order than AB. 

Let ACB, Fig. 43, be a right angle inscribed in a semicircle. 
Let BD be a tangent line, and let CE be perpendicular to BD. If 



92 



CALCULUS 



[§59 



the angle CAB approaches zero, BC, CD, BE, ED, CE, and arc 
BC are infinitesimals. From similar triangles 



AB BC BD 




AC~ BE~ BC' 

AB 
Since lim a^— 1, it follows that 


,. BC .. BD ^ 


Hence BC and BE, and BD and BC are infinitesimals of the same 


order. 


Again, 

BC CE CD 
AB BC ~ BD' 




D 

E 



,• BG ^ 

Smct! lim -ir-f, = 0, 

AB ' 



Fig. 43. 






Hence CE is an infinitesimal of higher order than BC, and CD 
is an infinitesimal of higher order than BD. 
Again, 

CB ^CE^ _CD 

AC ~ BE ~ CB 

Since lim -j-^ = 0, 

AC ' 

hm^ = lim^ = 0. 

Hence CE is an infinitesimal of higher order than BE, and CD is 
an infinitesimal of higher order than CB. 



§60] INFINITESIMALS 93 

Exercises 

1. Show that X — 2x^ and 3a; + x^ are infinitesimals of the same 
order as x approaches zero. 

2. Show that 1 — sin d and cos'' 9 are infinitesimals of the same 
order as 6 approaches g- 

3. Show that 1 — sin is an infinitesimal of higher order than 
cos as 6 approaches ^^ 

4. Show that sec a — tan a is an infinitesimal as a approaches -• 

5. Show that 1 — sin a is an infinitesimal of higher order than 
sec a — tan a as a approaches ^• 

6. Show that 1 — cos 6 is an infinitesimal of the same order as 0^ 
as 6 approaches zero. 

T ou +1, + lim sin g - g 

7. Show that ^^q = 0. 

o ou +1. + lim sine - 9 

8. Show that „ . „ — . — r— = 0. 

»=o sin 9 

9. Show that i™ *^^^T^^ = °- 

10. Show that 1™*^J^^ = 0. 

*=o tan d 

^^ n, . > , lim sin a — tan a 

11. Show that .^ 7 = 0. 

«=0 tan a 

12. Show that^'"^ /^".-^^"" = 0. 

«=o sin a 

60. Theorem. The limit of the quotient of two infinitesimals, 
a and /3, is not altered if they are replaced by two other infinitesimals, 

a B 

y and 5, respectively, such that lim -- = 1 and lim -: = 1. 



Proof : 



a 

y ~ 

a _ 7 

~B~ TT' 

lim — 

lim ^ = a lim -i = lim j i 

hm g^ 



94 CALCULUS [§60 

since 

lim — = lim -v = 1- 
7 

It is evident from the proof that the limit of the quotient is 

unaltered if only one of the infinitesimals, say a, is replaced by 

T 
another infinitesimal 7, such that lim - = 1. 

a 

Illustrations. 

L Since 

lim sin a _ 

a=0 ~ ~ ^» 



a 

lim 1 ~" cos a _ Hm 1 ~ cos a 
«-o sin a ~ "=o q. 



= 0. 



2. Since 



lim tan a 



a = 



a 



1, 



lim 1 — cos « _ lim 1 — cos « _ 
«-0 tan a "-0 a 

3. In Fig. 40, 

lim ^ ^ lim C;4 _ Urn C'A ^ 
cc^oAB »=OBC "=0AT 



Exercises 
l.Showthat^^"l ^"7" = i 
Hint. 1 — cos a = 2 sin^ ^- 

2. Show that lim ^i^«a--cos«) ^ 

a=0 a^ ^ 

3. Show that lim («-5)-sina ^ ^5. 

a=0 a 

4. Show that"'" ^"'"'« ' 



«=Ocos a sin- a ^* 
K «, ,, . lim 3 3:" - 4x3 

5. Show that ;,^0 2 x^-5x< = ^' 

Hint. Replace numerator by 3x* and denominator by 2x*. 
A .A A 

6. Show that "."^^—4^= ^^-^ = 2. 

X^ X^ X^ 



1611 



DIFFERENTIALS 



95 



61. Diflferentials. Let PT, Fig. 44, be a tangent line drawn 
to the curve y = f{x) at the point P. Let DE = Ax, RQ = 
A?/, and let angle RPT = t. 

From the figure, 

RM . .,. . 

= tanr = / (x), 



Ax 



or 



RM = fix) 'Ax. 

This is the increment which the function would take on if 
it were to change uniformly at a 
rate equal to that which it had 
at P. 

This quantity, j'{x)Ax, is 
called differential y, and is de- 
noted by dy. Its defining equa- 
tion is 



dy = f'{x)Ax. 



(1) 




Ax, the increment of the independent variable, is called differential 
X and is denoted by dx, i.e., Ax = dx. Equation (1) becomes 



dy = j'{x)dx\ 



(2) 



In Fig. 44, RM = dy and DE = PR = dx. 

In general, dy is not equal to Ay, the difference being MQ, 

Fig. 44. However, it will be shown that ^^^g ^ — ~ ^• 

lim RQ_f,(^s 



or 



lim \RQ_ Rm = .>r^. 

ix=0 RM PR J ^•^^' 



^^^0 IRM PRJ 



(3) 



RM . 



But, since p^ is constant and equal to/'(x), equation (3) becomes 
lim RQ 

Ax=0 RM 



1, 



> In the expression (2) for the differential of the function /(i), the first derivative 
is the coefficient of the differential of the argument, and for this reason it is sometimes 
called the differential coefficient. 



96 CALCULUS (§62 

at 

It is to be noted that dx is an arbitrary increment and that dy 
is then determined by this increment and the value of the deriva- 
tive, i.e., by the slope of the tangent at the point for which the 
differential is computed, dx and dy are then definite quantities 
and we can perform on them any algebraic operation. Thus we 
can divide (2) by dx and obtain 

t = f'M. (5) 

where dy and dx denote the differentials of y and x, respectively. 
Thus from the definition of differentials the first derivative may 
be regarded as the quotient of the differential of y by the differ- 
ential of X. 

It is to be observed, however, that this statement gives no new 
meaning to the derivative, since the derivative was used in the 
definition of the differential. 

62. Formulas for the Differentials of Functions. In accordance 
with equation (5) of the preceding section, any formula involving 
first derivatives can be regarded as a formula in which each first 
derivative is replaced by the quotient of the corresponding 
differentials. Thus, 

, /m\ du dv 



dx dx 



dx 



Each derivative being considered as a fraction whose denominator 
is dx, we can multiply by dx, and obtain 

vdu — udv 



e) = 



v 



In words, the differential of a fraction is equal to the denominator 
times the differential of the numerator minus the numerator times the 
differential of the denominator, all divided by the square of the denomi- 
nator. It will be noted that the wording is the same as that 
for the derivative of a fraction except that throughout the word 
differential replaces the word derivative. 



§62] DIFFERENTIALS 97 

The other formulas for derivatives which have been de- 
veloped are expressed below with the corresponding formulas for 
differentials. 







Formulas 


1. 


dc 


dc = 0. 


2. 


d{cu) du 
dx ~ dx 


d{cu) = cdu. 


3. 


d{u + v) du dv 
dx ~ dx dx 


d{u + v) = du + dv. 


4. 


du'* , du 
dx dx 


dw* = nM"~^ du. 


6. 


d(uv) dv du 
dx dx dx 


d(uv) = udv -{- vdu. 


6. 


, /u\ du dv 

a ) v~j wt— 

\v 1 dx dx 

dx ~ v^ 


, /u\ vdu — udv 


7. 


\v / dx 

dx ~ v^ 


, /c \ cdv 


8. 


dx v*^ 


/ c \ cndv 


9. 


du 
du dx 


dJ - ^" . 
2u^ 



The formula for the differential oi y = cu" can be put in the fol- 
lowing convenient form : 

(fy du 

10. — = n — ■} 

y u 

which is obtained directly by dividing dy = cnu'*~^du hy y = cW. 

The process of finding either the derivative or the differential of a 
function is called differentiation. 

The process of finding a function when its derivative or differential 
is given is called integration. 

We have no symbol representing integration when applied to 
derivatives. The symbol for integration when applied to dif- 



98 CALCULUS [§62 

forentials is j . Thus I Sx^dx = x^ -\- C. The origin of this 
symbol will be explained later. It is read "integral of." 

Illustrations. 

1. U y = Vn^S 

dy = Ui - x2)-5( - 2xdx) 
_ xdx 

Vl - x^' 
By formula 10, where w = 1 — x^, 

dy _ 1 — 2x dx 
J ~2 (1 - a;*) 

xdx 



2. If 2/ = 



1 -a;* 

X 



X2- 1 



^y = (^^^^Tp 

_ (x' — 1) c?a; — X (2a; dx) 
~ (x2_ 1)2 ■ 

^ _ (x^ + l)dx 
(x* - ly ' 
3. If d?/ = xdx, 



y = ixdx 

= A r2xdx 



=i-+c. 



4. If dy = xVl — x^ dx, 

y = Cx{l — x^)^dx 
= -H/HI -x2)2( _2xdx) 
(l-x^)^ 



+ C. 



§62] DIFFERENTIALS • 99 

dtj dx 

r' 

2/ = C(x - 1). 



5. If - , 

y X - 1 



by formula 10. 
ft If d]i xdx 



y x^ — 1 



dy _ 1 2xdx 
y ^ 2 x2- 1 

y = C\/a;2 - 1. 



Exercises 

Find dy in the following ten exercises : 

1. J/ = x2 - 3x - 2. ^ V: 

6. y = 



2. 2/ = 



(x-l)» 



X- 1 7. J/ = (x - l)(x» - 1)2. 

3. y = x^- x~^ - 3a;. 8. y = {x* + x - 2)». 



,_i 



4. y = (x - 2)*. 9. y = (x - 1)" 

6. y = (x2 - 2)i 10. y = (x2 - 1)~^. 

Integrate the following: 

11. fxMx. 

12. C{x^ - l)xdx. 

13. fCx' - 3x + 5) (x2 - 1) dx. 

14. r(x» - 2x - 6)3 (x - 1) dx. 

/dx 

16. Cs/x dx. 

"■ /t 

18. Cx^dx. 



100 



CALCULUS 



m 



19. 


dy 

y 


= 


X 


20. 


dy 


_ 


xdx 



x2- 1 



21. 




dy _ (x^ - 2x + 4)dx 
y ~ a;3-3x2+12a;-2' 



lim 

Ax 



Since^ 



63. Differential of Length of 
Arc : Rectangular Coordinates. 

Let PR, Fig. 45, = Ax, RQ = Ay, 
the chord PQ = Ac, and the arc 
PQ = As. (s represents the length 
of arc measured from some point 
A.) PT is the tangent at P. 

(Ax)2 + {Ayy 

m /^\2 _ 1 , lim (^Y__ 1 , (f^vy 
^0 \Ax) ^ "^ ^^=^0 \^xj ~ ^ "^ \dxl 

lim Ac 



A2:=0 ^s 



= 1, 



> When Ai is taken so small that the curve has no point of inflection between P 
and Q, the chord PQ < arc PQ < PT + TQ, or Ac < As < Pr + TQ. Whence, 



Therefore 



Then from (1), 



i<^<^+m 



(ptX 2 ^ (da:) 2 + (dy)2 ^ [^ \ dx) 
[ac J (Ax) 2 + (A2/)2 J _|_ /^\f 



lim /• 
Ax=0 I 



Ac I 



= 1. 



lim TQ 
Ax=0 -^ 



lim Aj/ — dy Ay 

^^-0 A^ a;^ 



r lim /. ''i/^l r 'im ^ l „ n 
= [Ax^oV^ - Ayjj [ax=0 Ac J "' 



lim ^ =, 
Ax— Aj/ 

lim As 
Ac 



Aa;=s=0 



(1) 



DIFFERENTIALS 101 

Ac can be replaced by As ( §60) . 

lira /As\ 2 ^ , , /M ' 
^=0 \Ax) "^ [dxj ' 



or 



(ds)2 = (dx)2 + (dy)2 (1) 



ds=Vl+(g)'dx (2) 



ds = Vi + a~)^<iy 



Vdy/ - (3) 

Equation (1) shows that the line PT, Fig. 45. represents ds. 
If T denotes the angle made by the line PT with the positive 
X-axis, 

dx = cos T ds 

dy = sin r ds. 

Illustration. Find the length of the curve y = fx^ between 
the points whose abscissas are 3 and 8. 

^= x^ 
dx 



(I)'- 



Substituting in formula (2), 



ds = Vl + X dx. 
Integrating, 

s= 1(1 + x)' +C. 
When X = 3, s = 0. Hence C = - Y, and 



s= iil + xy -W 

This formula gives the length of the curve measured from the 
point whose abscissa is 3 to the point whose abscissa is x. On 
placing X = 8 we obtain s = ^3^, the length of the curve from 



102 



CALCULUS 



I §64 



the point corresponding to x = 3 to the point corresponding to 
X = 8. 

Exercises 
Find the differentials of the length of the following curves: 



1. y = x'. 

2. x^ + y^ = 4. 

3. 2/ = xK 

4. 2/2 = X. 



5. 3x2 + 4y2 = 12. 

6. xy = 1. 

7. xj/2 = 1. 

8. 7/ = x-^ 



64. The Limit of 2f(x)Ax. Let y = /(x) be a continuous 
function between x = a and x = 6. In §50 it was shown that 
the area bounded by the curve, the X-axis, and the ordinates 
X = a and x = 6 is given by the formula 

A = F{b) - F{a), (1) 

where F{x) = J f(x)dx. A second expression will now be found 
for the area. Divide the interval h — a, Fig. 46, into n equal parts 
and at each point of division erect an ordinate. Complete the 
rectangles as indicated in the figure. 



!/=/(«) 




The sum of the rectangles of which DEQ'P is a type, is approxi- 
mately equal to the area ABUV. The greater n, the number of 
rectangles, i.e., the smaller Ax, the closer will the sum of the 
rectangles approximate the area ABUV. We say then that 



_ lim 



1"^ 2 DEQ'P, 



§65] DIFFERENTIALS 103 

or 

x = b 



The above expression (2) represents the actual area and not 
an approximation to it, as can be shown by finding the greatest 
possible error corresponding to a given number of rectangles and 
then proving that this error approaches zero as the number 
of rectangles becomes infinite. Thus it is easily seen that the 
difference between the true area A and the sum of the rec- 
tangles is less than the area of the rectangle RSTU. The altitude, 
f{b) — f{a), of this rectangle is constant while the length of the 
base, Ax, approaches zero. Hence the area of RSTU approaches 
zero. Therefore the limit of the sum of the rectangles is the 
area sought. 

On equating the two expressions for A, given by (1) and (2), 
we have 

x = b 



where 



Fix) = ff(x)dx. 



This equation is the important result of this section. It gives a 
means of calculating 



Ax= 

X =a 



For, to calculate this limit we need only to find the integral of 
f{x)dx and take the difference between the values of this integral 
at X = a and x = h. The result of this section will be restated 
and emphasized in the next section. 
65. Definite Integral. The expression 

x = a 

which was introduced in the preceding section is of such great 



104 CALCULUS [§65 

importance that it is given a name, "the definite integral of f{x) 
between the limits a and b," and is denoted by the symbol 



i 



f{x)dx. 



Equation (3), §64, gives a means of calculating the value of the 
definite integral. 

The function F(x), the integral oi f{x)dx, is called the indefinite 
integral of j{x)dx in order to distinguish it from the definite inte- 
gral which is defined independently of it, viz., as the limit of a 
certain sum. 

We have then the following definition and theorem: 

Definition. Let J{x) be a continuous junction in the interval from 
X = a to X = b, and let this interval be divided into n equal parts of 
length Ax by points Xi, Xz, Xi, . . ., x„_i. The "definite integral of 
f{x) between the limits a and b" is the limit of the sum of the products 
f{xi) Ax formed for all of the points Xq = a, Xi, xj, . . . , x„_i, as the 
number of divisions becomes infinite. 

Theorem. The definite integral of f(x) between the limits a and 
b is calculated by finding the indefinite integral, F(x), of f{x)dx and 
forming the difference F{b) — F{a). 

The symbol for the definite integral, 

fix)dx, 



r 



is read "the integral from a to b of f{x)dx." As we have seen, it 
means 

h 



Jf{x)dx = J2i Zff{x)Ax. 
a a 



Many problems, such as finding the work done by a variable 
force, the volume of a solid, the coordinates of the center of 
gravity, lead to definite integrals. But, no matter how a 
definite integral may have been obtained and no matter what 
other meaning it may have, it can always be regarded as repre- 
senting the area included by the curve y = f(x), the X-axis, and 
the ordinates x = a and x = b, provided that /(x) is a function 



DIFFERENTIALS 105 

which can be represented by a continuous curve. This fact, that 

f{x)dx 



r 



can be regarded as representing an area, enables us to calculate 
its value. For the area in question is equal to F{b) — F{a), 
where F{x) is the indefinite integral of f{x)dx. Consequently we 
have, in all cases, 

f{x)dx = F{b) - F{a). 



r 



This is often written 



X 



b 

j{x)dx = F{x) 



= F{b) - Fia), 



to show how the result is to be calculated. Thus 



I 



x^dx = -^ 



2^ _ P _ 7 
3 3 ~ 3' 



Exercises 

Evaluate the following definite integrals: 

r C'dx C 

1. I (2x + Z)dx. 2. I ^- 3. I Va^ + x" xdx. 

Ji Ji Jo 

66. Duhamel's Theorem. If ai, ocz, as, • ■ •, oin are n in- 
finitesimals of like sign, the limit of whose sum is finite as n becomes 
infinite, and if /3i, jSz, /Ss, • • , /3„ are a second set of infinitesimals 
such that 

lim §J _ 1 
n=co ^. - A. 

where i = 1, 2, 3, • • • , n, then 

lim V „.. = lim V p,. 

n=m ^ n=co ^ t^*' 

1 1 

Proof. Let — = 1 + €<. 

OCi 



106 CALCULUS [§66 

Since 

lim ^i _ . 
n=co „. A. 

^^ €i = 0. 

n=oo 

At first let it be assumed that the a's are positive. Let E be the 
numerical value of the largest e, i.e., 

E >. le<', i = 1, 2, 3, • • • , n. 
Then, since /3i = «» + e.a,, i = 1, 2, 3, • • • , n, 
ai — Eai < ^i £ ai + Eai 
az — Ea-i <. ^1 ^ a2 + Ea^ 

an - EUn <^n<an-\- E Un- 

Adding, we get 

i = n x=n »' = » 

{l-E)%ai<^^i<{l-\-E) X cii. 
1=1 i = 1 »■ = 1 

^^E = Q, 

n=oo 

n i = n 

lim ^ a< = lim ^ /3i 
t = 1 » = 1 

and the theorem is proved. 

If the a!s are negative, it will be necessary to change the proof 
just given, only by reversing the signs of inequality. 

Section 64 furnishes an illustration of this theorem. In this 
example the limit of the sum of the infinitesimal trapezoidal 
areas DEQP is finite as n becomes infinite, since it is the area 
sought. 

DEQ'P < DEQP < DEQP', 
(see Fig. 46), or 

yAx < DEQP < {y + Ay) Ax, 
or 

DEQP y + Ay _ 
^ ^ DEQ'P ^ y 

This shows that the limit of the ratio of the trapezoidal area to 



Since 



n=< 
t = n 



§67] DIFFERENTIALS 107 

the area of the corresponding rectangle is 1 as n becomes infinite. 
Then by Duhamel's Theorem, 

Um 2 DEQ'P = ^^^ 2 DEQP = A, 

Since we are able to replace the infinitesimals DEQP by the 
infinitesimals DEQ'P, we may calculate the area which is the sum 
of these infinitesimals by means of the definite integral. This 
is a characteristic process in the use of the definite integral. 
The quantity sought is subdivided into n portions which are 
infinitesimals as n becomes infinite. These are replaced by n 
other infinitesimals of the form f{xi) Ax. The limit of the sum of 
the latter infinitesimals is a definite integral. 

Since the limits of the two sums are equal by Duhamel's 
Theorem, the definite integral is equal to the quantity sought. 

Illustrations of the applications of Duhamel's Theorem to 
obtain definite integrals representing work, force, volume, etc., 
follow. 

67. Work Done by a Variable Force. In §51 there was found 
the work done by a variable force, /(s), in producing a displacement 



Fig. 47. 

from s = a to s = &. We shall now obtain the same result by 
building up the definite integral which represents the work. 
Divide the total displacement h — a, Fig. 47, into n equal parts 
of length As. The force acting at the left end of one of these 
parts is /(s), while that acting at the right end is /(s + As). 
The total work done in producing the displacement, 6 — a, is 
approximately 

8 = 6 

s = a 

The actual work is the limit of this sum as As approaches zero.^ 

' This step can be justified by using Duhamel's Theorem. Let Au) represent the 
work done in producing the displacement As. Then 



108 CALCULUS [§67 

Illustration 1. The solution of the problem of Illustration 1, 
§51, is expressed by 

4 



I 



SOsds = 15s2 



= 180. 

2 



Illustration 2. The solution of the problem of Illustration 2, 
§51, is expressed by 



w = 



kmM I — - = — kmM — = — kmM \r I 

J„ s- s „ L6 a J 



= kmM\ 

Illustration 3. In solving the problem of Illustration 3, §51, 
we can write 



_ lim 



« = o 






where t)2 and t^i are the volumes corresponding to s = a and s — b, 
respectively. Since pv'' = C, p = ^' and 



, C'dv _ C J 



C 



The student will complete the numerical work. 

n= 00 
But /(s)As < Au) < /(s + As)A8, 

, ^ Aw /(s-fAs) 
°' ^</(.)A«<-7(«) 

Then 

lim Am 



1 _ A; ^'"'~'' ~ '"'~''^- 



Hence by Duhatnel's Theorem 
b 



As=0 /(s)A8 



So 2^-= ii"o 2/WA.=/j'/(,M.. 

o a " 

s = b c 

s = a *f "■ 



DIFFERENTIALS 



109 



Exercises 




1. Set up and evaluate definite integrals representing the work 
sought in Exercises 1-5, §51, Chapter VI. 

2. Water is pumped from a round cistern whose median section 
is a parabola. The cistern has a diameter of 8 feet at the top and 
it is 16 feet deep. The water is 10 feet deep. Find the work done 
in pumping the water from the cistern if the discharge of the pump 
is 3 feet above the top of the cistern and if the friction in the pump 
and the friction of the water in the pipes are neglected. 

3. Find the work done by a gas in expanding in accordance with 
the law pv^-* = C from a volume of 10 
cubic feet to one of 12 cubic feet, if 
when V = 9 cubic feet p = 100 pounds 
per square inch. \B 

4. Find the work done in stretching 
a spring whose original length was 15 
inches from a length of 16 inches to a 
length of 18 inches if a force of 40 
pounds is required to stretch it to a 
length of 16 inches. 

6. Find the work done in compress- 
ing a spring of original length 5 inches 
to a length of 3§ inches, if a force of 
900 pounds is required to compress it 
to a length of 4 inches. 

6. The force due to friction is pro- 
portional to the component of force Fig. 48. 
normal to the surface over which a body 

is being moved. Find work done in dragging a body weighing 100 
pounds from the base to the top of a slide in the form of a segment 
of a sphere, Fig. 48, if the distance AB = 200 feet and the radius of 
the sphere is 500 feet. Express the result in terms of /*, the coeffi- 
cient of friction. 

68. Volume of a Solid of Revolution. The area bounded by 
the curve y = f{x), Fig. 46, the ordinates x = a and x = b, 
and the X-axis, is revolved about the X-axis. Find the volume of 
the solid generated. 

Divide the interval AB = 6 — a on the X-axis into n equal parts 
of length Ax and pass planes through the points of division 
perpendicular to the X-axis. These planes divide the volume into 



H 



110 CALCULUS [§68 

n portions, Ay. A typical portion can be regarded as generated 
by revolving DEQP, Fig. 46, about the base DE in the X-axis. 
Replace the volume of this slice by that of the cylinder generated 
by the revolution of DEQ'P about the Z-axis. Its volume is 
TTj/'Ax. The total volume is then 



or 



-r. 



Illustration. Find the volume between the planes x = 1 and 
X = 3 of the solid generated by revolving the curve y = x^ -\- x 
about the X-axis. 

/»3 /»3 

F = TT I {X^ + Xydx = TT \ {x* + 2x3 _|_ x^)(lX 

= X[U' -I- \X* -I- W]\ = TTXV^X^ + hx + ii; 

= 27ir(| -1-3-^1)- x(i + i 4- i) = -^ tPr. 

Exercises 

1. Find the volume between the planes x = and x = 3 of the solid 
generated by revolving the parabola y^ = 6x about the X-axis. 

2. Find the volume of a sphere of radius r. 

3. Find the volume of the ellipsoid of revolution generated by 
revolving the ellipse 

16 "^ 9 

about the X-axis; about the K-axis. 

4. Find.the volume between the planes x = and x = 4 of the solid 
generated by revolving y^ = x^ about the X-axis. 

6. Find the volume of the solid generated by revolving x' + y^ = a' 
about the X-axis. 

6. Find the volume generated by revolving y^ = 2ax — x* about 
the X-axis. 

7. Find the volume generated by revolving the oval of 
y* = x(x — l)(x — 2) about the X-axis. 



§69] 



DIFFERENTIALS 



111 



69. Length of Arc: Rectangular Coordinates. In §63 the 

length of arc of a curve was found by integrating its differential. 
We shall now express the length of arc by means of a definite 
integral. 

To find the length of arc APQB, Fig. 49, divide CH into n 
equal parts of length Ax each. At the points of division erect 
ordinates dividing the arc AB into n parts of which PQ is one. 
The length of arc AB is defined by 



g ^ lim V ^ 

n=oo ^^ 



3 


P Q 






A 

/ 


y 


^ 




AC 




\ 


B 

\ 


O 


C D E 

■«-o-*j 


H- 



















Fig. 49. 
where Ac is the length of the chord PQ. Then 
s = ^^^ y,V(Ax)^+(Ayy 



Since 






= 1, 



it follows by Duhamel's Theorem that 

b 

lim 
s 



=:iT.x^^uM'^ 



112 CALCULUS [§70 

Hence, 

« = I \/l + l^) dx. 



r>/^^ 



Exercises 

l.^Find the length of the curve y = x^ between the points (0, 0) 

and'(l, 1). 

z 1 1 

2. Find the entire length ofx* +y* = a^- 

3. Find the entire length of x^ + y^ = a^. 

4. Find the length of y^ = 4:X^ between the points (0, 0) and (4, 16). 
70. Area of a Surface of Revolution. The portion AB, Fig. 49, 

of the curve y = f{x), between the ordinates x = a and x = b, 
is revolved about the X-axis. Find the area, S, of the surface 
generated. 

Pass planes as in §69 perpendicular to the X-axis through the 
equidistant points of division of the interval CH = h — a. 
Denote the convex surface of the frustum of the cone generated by 
the revolution of DEPQ by AF. The area, S, of the surface of re- 
volution will be defined as the limit of the sum of the convex 
surfaces, AF, of these frusta as n becomes infinite, i.e., as Ax 
approaches zero. Then, 



„ lim V A p lim V o ^ + ('^ + ^V^ 



Ac 



Ay 
By Duhamel's Theorem we can replace y + -„" by y, since 



lira 2/ + |Aj/ 
Ax-o V, = 1- Hence, 



Since 






lim , — , 

\x=0 I T — = 1> 

'dy\i 



§71] DIFFERENTIALS 113 



Therefore 



a 

yds, 

x — a 



5 = 2x I v\il + [~\''dx 



= 27r 

where ds is the differential of the length of arc. The latter form 
is easily remembered since 2Tryds is the area of the strip of surface 
generated by revolving ds, the differential of arc, about the X-axis 
at a distance y from it. If it is more convenient to integrate with 
respect to y, ds can be replaced by 



v^d)^"- 



and the limits are the values of y corresponding to x = a and x = b. 
Thus 

Wl + U) dy = 2ir\ yds. 

Exercises 

1. Find the surface between the planes x = and x = 5 of the 
paraboloid of revolution obtained by revolving t/'* = 4x about the 
X-axis. 

2. Find the surface of the sphere generated by revolving x'' + 2/^ =^* 
about the X-axis. 

3. Find the surface of the right circular cone whose altitude is 10 
feet and the radius of whose base is 5 feet. 

4. Find the surface of the solid generated by the revolution of 

2 2 2 

a^ + 2/' = a* about the X-axis. 

71. Element of Integration. The first step in setting up a 
definite integral is to break up the area, volume, work, length, or 
whatever it is desired to calculate, into convenient parts which 
are infinitesimals as their number approaches infinity. These 
parts are then replaced by other infinitesimals of the typical 

8 



114 



CALCULUS 



[§72 



form fix)dx, which must be so chosen that the limit of the ratio 
of each infinitesimal of the second set to the corresponding 
infinitesimal of the first set is one. fix)dx is called the "element" 
of the integral or of the quantity which the integral represents. 
Thus the element of volume is wy^dx, that of area is ydx, that of 
work is Fdx. 

If the magnitude which it is desired to calculate is broken up into 
suitable parts, the expressions for the elements can be written 
down at once. The best way of retaining in mind the formulas 
of §§68, 69, and 70 is to understand thoroughly how the elements 
are chosen. The process of writing down the element of integra- 
tion at once becomes almost an intuitive one. 

72. Water Pressure. The pressure at any given point in a 
liquid at rest is equal in all directions. The pressure per unit 

area at a given depth is equal to the 
pressure on a horizontal surface of 
unit area at that depth, i.e., to the 
weight of the column of liquid sup- 
ported by this surface. This weight 
is proportional to the depth. Hence 
the pressure at a depth x below the 
surface of the liquid is given by the 
formula p = kx. If the liquid is 
water and the depth x is expressed 
in feet, k = 62.5 pounds per cubic 
foot. 
The method to be used in finding the water pressure on any 
vertical surface is illustrated in the solution of the following 
problems: 

1. Find the pressure on one side of a gate in the shape of an 
isosceles triangle whose base is 6 feet and whose altitude is 5 feet, 
if it is immersed vertically in water with its vertex down and its 
base 4 feet below the surface of the water. 

Take the origin at the vertex of the triangle, the axis of x vertical, 
and the axis of y horizontal, as in Fig. 50. The altitude is sup- 
posed to be divided into n equal parts and through the points of 
division horizontal lines are supposed to be drawn dividing the 
surface into strips. The trapezoid KHMN = AA is a typical 




§72] DIFFERENTIALS 115 

strip. Denote the pressure on this strip by AP. The abscissa 
of the lower edge of the strip is x and the pressure at this lower 
edge is A;(9 — x). Then the total pressure is 

P = n^^Xm-x)LA. (1) 

In accordance with Duhamel's Theorem we can replace AA 
by 2i/Ax. 

x = 



or 



Since 



= 2A; I (9 - 



x)ydx. (3) 



V = 



3x 
V 



•5 

P = ~ \ (,9-x)xdx 



_6kr 

5 Jo 

= g- "2- - 3^ = 5312.5 pounds. (4) 

In general, if u denotes the depth below the surface of the liquid 
and z denotes the width, at the depth u, of the vertical surface 
on which the pressure is to be computed, 

P = k \ uzdu, (5) 

where a and b are the depths of the highest and lowest points, 
respectively, of the surface. For, 

p = lim X\kuAA = lin^ X kuzAu = k \ uz du. 

n^a, ^^ Au=0 •^T I 

2. Find the total pressure on a vertical semi-elliptical gate 
whose major axis lies in the surface of the water, given that the 
semi-axes of the ellipse are 8 feet and 6 feet. Take the origin at 



116 



CALCULUS 



[§73 



the center, the axis of x horizontal and the axis of y positive down- 
ward. The element of pressure is 



and the total pressure is 



2kyx dy 



^ = 4 



yxdy. 



X is expressed in terms of y by means of the equation of the 
ellipse, 



Then 



— -u ?^ = 1 
64 "^ 36 



P = 2ki I yVSQ - y^ dy. 



Exercises 

1. Find the pressure on the vertical parabolic gate, Fig. 51: (o) 
if the edge AB lies in the surface of the water; (b) if the edge AB lies 
5 feet below the surface. 




2. Find the pressure on a vertical semicircular gate whose diameter, 
10 feet long, lies in the surface of the water. 

73. Arithmetic Mean. The arithmetic mean, ^, of a series of 
n numbers, Oi, 02, aa, • • • , a„, is defined by the equation 



or 



nA = ai + a2 + as + • • • + a„, 
ai + a2 + as + • • • + a„ 



A = 



That is, A is such a number that if each number in the sum 



§741 DIFFERENTIALS 117 

Oi + c[2 + cfs + * • ' -\- ttn be replaced by it, this sum is 
unaltered. 

74. Mean Value of a Function. We can extend the idea in- 
volved in the arithmetic mean to other problems. 

Illustration 1. Suppose a body moves with uniform velocity 
a distance of 1 foot during the first second, a distance of 2 feet 
during the second second, a distance of 3 feet during the third 
second, and so on for 10 seconds. At the end of 10 seconds the 
body would have moved 1 + 2 + 3+ • • -+10 = 55 feet. 
The mean, or average, velocity of the body is the constant velocity 
with which the body would describe this distance in the same 
time. It is equal to 5.5 feet per second. 

If the velocity of the body instead of changing abruptly as 
indicated above were changing continuously in accordance with 
the law V = t, the total distance s traversed in 10 seconds would be 

•10 

tdt = 50. 



J "10 /*1 

vdt = j 
Jo 



From this equation 



The mean velocity, V, the constant velocity which a body must 
have in order to traverse the same distance in the same length 
of time, is 50 -i- 10 = 5 feet per second. This can be expressed 
by the formula 

J "10 /»10 

Vdt = I vdt. 
Jo 

vdt 

10 

In general, if i; = j{t), the mean velocity, V, of the body in 
the interval of time between t = a and f = 6 is expressed by the 
equation 

= { At)dt, 



or, since F is a constant, 

S{t)dt 
V = 



Vdt= I 

a Ja 

Ja 



b — a 



118 CALCULUS [§74 

V is the constant velocity, which replacing the variable velocity, 
" = /(O, at every instant in the interval between t = a and t = b, 

gives the same distance traversed, i.e., leaves the value of the 

nb 
integral, I v dt, unchanged. 



Illustration 2. Consider the work done by a variable force / 
acting in a straight hne, the X-axis, and producing a displacement 
from a; = a to X = 6. If the law of the force is f = (f){x), the 
mean force F in the interval from a; = a to x = 6, or the constant 
force which would do the same work while producing the dis- 
placement 6 — o, is given by the equation 



nb f*b 

I Fdx = I 0(x)dx, 

f. 



or 

(l){x)dx 

F = 

b — a 

F is a constant such that if, in the integral I 4>{x)dx, the func- 
tion ^(x) be replaced by it, the value of the integral remains 
unchanged. 

Illustration 3. Let a unit of mass be situated at each of the 
points on the X-axis whose abscissas are Xi, X2, X3, • • •, x„. 
The X-axis is taken horizontal and the masses are acted upon 
by gravity. We shall find the distance, x, from the origin at 
which the n masses must be concentrated in order that the sum of 
the moment about the origin of the forces acting on the masses 
shall be unchanged. 

Clearly x must satisfy the equation 

gnx = g{xi + X2 + X3 -{- • • • + x„), 
or 

Xi + X24-X3+ • • . +x„ 

X = • 

n 

If there are mi, mj, ms, • « • , 7n„ units of mass concentrated 
at Xi, X2, Xs, • • • , x„, respectively, the mean moment arm, x, 
the distance from the origin at which the masses must be con- 



§74] DIFFERENTIALS 119 

centrated in order that the sum of the moments about the origin 
of the forces acting on the masses shall be unchanged, is given by 
the equation 

(mi + m2 + • • • m„)x = niiXi + W2X2 + • • • + m„x„, 
or 

- 2 THiXi 



2wi 



-, (i= 1,2,3,. . .,n). (1) 



rr is a constant such that if in the sum 2 m,x, each of the num- 
bers Xi, X2, • • • , x„ be replaced by x this sum is not changed. 

Now let there be a continuous distribution of matter along the 
X-axis from x = a to x = b. Divide the interval b — a into n 
segments each of length Ax. An expression for the approximate 
sum of the moments about the origin of the forces acting on the 
mass is 2 gxAmt, where Am, is the mass of the segment AZf. 
An expression for the approximate force is 2 ^Am,-. Hence an 

expression for the approximate x is „ . — -*• It is readily seen 

that as Ax approaches zero, the numerator approaches the total 
moment and the denominator approaches the total mass. Hence 



lim ^ ^ T^ 

Am=o ZigAm I dm 



(2) 



X is a constant such that if in the integral, I x dm, x is replaced 



, I xdm, 



by X, the value of the integral is unchanged. 

For example, if the density is proportional to x^, i.e., is equal to 
kx^, the element of mass, dm, is kxHx, and we have 



J'*b nb 

xkxHx I x^dx 

_ a _ Ja 



•6 Ch 

Mx 

3 b* - a\ 

»b 7'6 4 fe» — a' 



kx^dx I x^dx 

a Ja 

The mean value, M, cf the function fix) with respect to the 



120 CALCULUS [§74 

magnitvde u, which is a function of x, is defined by the equation 

(3) 

where M is a constant, or 



Jf*x = b r*x = b 

Mdu = I f{x)du, 
x = o Jx = a 

J'*x = b 
f{x)du 
x = a 



M = 



'1 = 6 

du 



J^di 
x = a 



(4) 



M is a constant such that if the function f{x) is replaced by it in 

Jr*x = b 
f{x)du, the valv£ of the integral is not changed. 
a; = o _ 

In (2), X is the mean value of x with respect to the magnitude, 
m. 

A particular case of (4) is that in which u = x. Then (4) 
becomes 

M = ,— ^^ I f (x) dx. (5) 



1 p 



Illustrations 1 and 2 are cases of 
this type. 

When w = a;, as in equation 
(5), M can be interpreted as 
the altitude, AC, of a rectangle 
wth base AB = b — a, Fig. 
52, whose area is equal to the 
area bounded by the curve y = 
■X f{x), the A'-axis, and the ordi- 
nates x = a and x = b. From 
this standpoint M is called the 
mean ordinate of the curve y = 

f{x) in the interval from x = a to x = b. 
Illustration 4. Find the mean ordinate of the curve y = x' 

between the ordinates a; = and x = 2. 




Fig. 52. 



M = ^-^ I f{x)dx = § I x^dx = ix3 



4. 



§74] DIFFERENTIALS 121 

Illustration 5. Find the mean with respect to w of x between 
the limits u = \ and tt = 9, if m = x^. 

J'*u =» 9 (*u = 9 

Mdu = I xdu, 
u = 1 Ju = 1 

J 3 /»3 

2xdx= I 2x»dla:, 

8M = V, 
M = V. 

Exercises 

Find the mean ordinates for the following curves: 

1. y = X* between x = and x = 3. 

2. 2/ = x^ between x = 2 and x = 4. 

3. 2/ = 3x3 between x = and x = 2. 

4. y = 3x3 between x = 1 and x = 3. 

5. J/ = x* between x = and x = 1. 

6. Find the radius of the right circular cylinder of altitude 3 whose 
volume is equal to the volume between the planes x = 2 and x = 5 of 
the solid generated by revolving 7/ = x + x'' about the X-axis. 

7. Find the radius of the right circular cylinder of altitude h — a 
whose volume is equal to the volume between the planes x = a and 
X = fe of the solid generated by revolving y = /(x) about the X-axis. 

8. The density cf a thin straight rod 10 inches long and of uniform 
cross section is proportional to the distance from one end. Find the 
mean density of the rod. 

9. Find the mean velocity of a freely falling body between the 
time ^ = 1 second and t = 3 seconds. 

10. The density of a rod is given by p = 3x2, 'v^here x is the distance 
from one end. Find the mean density if the rod is 10 inches long. 

11. Find the mean moment arm in the case of the rods of Exercises 
8 and 10, about a horizontal axis through the end of the rod (x = 0). 
The rods are horizontal, and perpendicular to the axis about which 
moments are taken. The rods are supposed to be acted upon by 
forces due to gravity alone. 

12. Find the mean ordinate of a semicircle, the ends of which are 
upon the X-axis. 



CHAPTER VIII 
CIRCULAR FUNCTIONS. INVERSE CIRCULAR FUNCTIONS 

Up to this point only functions have been discussed which are 
simple algebraic combinations of powers of the dependent variable. 
Many interesting applications of the calculus to the study of 
these functions have been given. We shall now take up the 
study of the appUcation of the methods of the calculus to another 
very important class of functions, the circular functions. It is 
apparent that the principles developed in the preceding chapters 
are equally applicable to the circular functions and to the 
algebraic functions. 

As the student has already learned, the circular functions occur 
very frequently in the study of the physical sciences and their 
applications, because by means of them periodic phenomena can 
be studied. 

76. Derivative of sin u. 
Let 

y = sin u. 
y -\- Ay = sin (u + Am), 

Ay = sin {u + Aw) — sin u 

— sin u cos Au + cos u sin Au — sin u, 
Ay _ cos w sin Am sinM(l — cosAm) 
Aw Am Am 

Then 

lim A^ _ lim sin Am . lim 1 - cos Am 

A«=o Am ~ ^°^ "^"=0 Am ®^° ^^«-o Am * 

Hence by §56 and §58 

dy 

^ = cosw.. (1) 

Whence 

dy du 

di = «°«^dx" (2) 

122 



§75] 



CIRCULAR FUNCTIONS 



123 



The corresponding formula for dy is 
dy = cos udu. 
It has thus been shown that 



d(smu) du 

— J = cos u j- 

dx dx 



(3) 
(4) 



and 



d(sinu) = cos udu. 

Well known properties of the function y = sin u can be verified 
by formula (1). Thus sin u is an increasing function between 

tt = and u = t), and between u — -^ and u = 27r, and decreas- 
ing between w = ^ and w = -p" The same facts are shown by 

re I I I p^ I I I I I I I I I I I 

G_u 




Fig. 53. 



the derivative, cos u, which is positive between w = and « = o' 



Stt 



and between u = -^ and u = 27r, and negative between u = ^ 



and u = 



Stt 



Further, sin u has maximum and minimum 

values for u = ^ and u = -^» respectively. The same facts 

are shown by the derivative, cos u, which becomes zero at these 

points and changes sign at ^ from plus to minus, and at -^ from 

minus to plus. 

The slope of the sine curve is approximately the slope of the 
diagonal PQ of a rectangle in Fig. 53. The greater the number of 
equal parts into which the circumference of the circle is divided 
and hence the smaller the subdivisions of the arc, the closer do the 
slopes of these diagonals approach the slopes of the tangents. 



124 CALCULUS [§76 

76. Derivatives of cos u, tan u, cot u, sec u, esc u. The 

derivatives of the remaining circular functions can be obtained 
from that of the sine. 
Let y = cos u. Then 



y = sin 



and 



dy /x X^(I-^) 
^ = cos^2-i.j- ^^^ 

du\ 



= cos (I - u) (- ^) 
du 





= — sm « -r' 
dx 


Hence 






d(cos u) . du 

— ^ = — sm u :i— 

dz dz 


and 






d(cos u) = — sin u du. 


By writing 






sin u 

tan u = » 

cos u 




cos u 

cot u = --. » 

sin u 



and 



sec u 



1 

CSC u = 



(1) 



sin u 
the student will show that 

d(tan u) du 

— ^ = sec^'u^ » or d(tan u) = sec^u du (2) 

d(C0t U) du Ar 4. \ 2 A f}\ 

— J = — csc'^u , » or d(cotu) = — csc^ u du (3) 

d(sec u) * du ^r \ *. a /^^ 

— -V = sec u tan u , ' or d(sec u) = sec u tan u du (4) 

d(cscu) . du ,. . , , ._. 

- — J = — CSC u cot u J » or d(csc u) = — csc u cot u du (5) 



§76] CIRCULAR FUNCTIONS 125 

Illustration 1. Find the first and second derivatives of 
3 sin (2a; - 5). 

rf[3sin (2a; -5)] _ d[sm (2x - 5)] 
dx dx 

d(2x - 5) 



dx 



= 3 cos (2a; - 5) 

= 6 cos (2a; — 5). 
Differentiating again, 

d'[3sin (2a; -5)] ^ rf[cos(2x-5)] 
dx^ dx 

= —6 sin (2x — 5) ^ 

= -12sin(2x - 5). 

dv 
Illustration 2. li y = sin 2a; cos x, find j— Since sin 2x cos x 

is the product of two functions, apply formula (1) §40. 

dy • „ / . ^ dx . , X / ^ X d2x 
-T- = sin 2a;( — sin x)-i — |- (cos x) (cos 2x) -r- 

= 2 cos x cos 2a; — sin x sin 2x. 

dv d^v 

Illustration 3. If ?/ = 3 sin x + 4 cos x, find --r- and -r-^' 

-p = 3 cos X — 4 sin X (6) 



From (6) 



d^y 

J— 2 = —(3 sin X + 4 cos x) = —y. (7) 

-r- = 4 cos X (f — tan x). 



TT 

When < x < ^, cos x is positive. The second factor, f — tan x, 

is positive when x < tan~^ (^), and negative when x > tan"^ (f ). 
Thus, when x is in the first quadrant the function has a maximum 
value coriiesponding to x = tan~i (|). 

When r, < a; < TT, -T- is negative. 

3x 
When TT < X < -^, cos x is negative, and J — tan x is negative 



126 



CALCULUS 



[§76 



when X < tan~^ (f), and positive when x > tan-' (f). Thus 
when X is in the third quadrant the function has a minimum 
value corresponding to the value x = tan~^ (j). 

When -^ < x< lir, -3- is positive. 

The same facts can be seen directly from the function, for it 
can be put in the form 

y = 5(5 cos X + t sin x). 



Let cos a = i and sin a 



Then 



or 



y = 5(cos x cos a + sin x sin a), 
y = 5 cos (x — a). 



In polar coordinates this represents a circle passing through the 

origin, with a diameter of 5. (See Fig. 54.) x is the vectorial 

angle and y the radius vector. The 

diameter OB makes an angle a with 

the polar axis. As x varies from 

to IT the circle is described, and as x 

varies from tt to 27r, y is negative 

and the circle is described a second 

time. 

Hence y has a maximum value 5 

when X is equal to a, and a mini- 

FiG. 54. mum value —5 when x is equal to 

a -\- IT. 

dy d^y 

Illustration 4. li y = tan' 3x = (tan ZxY, find -r- and -j-^* 

The function is of the form y = w. Hence 
g = 3(tan3x)^^^^^ 




= 3 tan^ 3x sec^ 3x 



d3x 
dx 



= 9 tan^ Zx sec^ 3x. 



d^ 
dx* 



9 tan* 3x 



c?(sec* 3x) 
dx 



+ sec* 3x 



(i(tan* 3x)- 
dx 



§76] CIRCULAR FUNCTIONS 127 



= 18 r 



tan2 3x-2 sec 3a; ^ , ' + sec^ 3x-2 tan 3x - ■ 



dx dx 

dSx 



tan^ 3a; sec 3a; sec 3a; tan 3a; , 

dx 

dSx^ 
+ sec'^ 3x tan 3a; sec^ 3x -i — 

= 54 (tan' 3x sec^ 3a; + tan 3x sec* 3a;) 
= 54 tan 3a; sec'' 3x(tan2 Sx + sec' 32). 

Illustration 5. If -r- = cos x, find y. 
dy 

-T- = COS x. 

dx 

y = sinx -\- C. 

dv 
Illustration 6. If -p = cos 3a;, find y. 

dy .r „ d3a;"| 
^=i[cos3x^J. 

The expression within the bracket is the derivative of sin 3x, 

hence, 

2/ = ^ sin 3x + C. 

dv 
Illustration 7. If ",^ = sin 3a;, find y. 



Hence 



= -i[-sin3a;^]. 



?/ = — ^ cos 3a; + C. 
Illustration 8. If -j- = sec^ 2x, find y. 

1/ = 5 tan 2a; + C. 

dv 
Illustration 9. If -^- = sec 5x tan 5x, find y. 



Hence 



= I sec 5x tan 5x -r— . 



128 CALCULUS [§76 

Hence 

y = 6 sec 5a; + C. 

Illustration 10. li dy = cos 3x dx, find y. 

y = \ cos Zxdx 
= \ \ cos 3x rf(3x) 
= \ sin 3x + C'. 
Illustration 11. li dy = sin'^ 2x cos 2x dx, find ?/. 

y = j (sin 2x)^ cos 2x dx 
= i 5/3 (sin 2x)2 cos 2x rf(2x) 

= i J3(sin 2xyd{sm 2x). 

Hence 

2/ = Ksin2x)« + C. 

Illustration 12. If d?/ = tan^ 5x sec^ 5x dx, find y. 
y = j tan^ 5x sec^ 5x dx 
= i i J 4 (tan 5x)^ sec'' 5x d(5x) 

= "sHr j 4 (tan 5x)' d(tan 5x). 

Hence 

y = ■2V(tan Sx)^ + C. 
Illustration 13. 
J sin 5x cos 3x dx = J Msin (5x + 3x) + sin (5x — 3x)]dx 

= 5 I sin 8x dx + ^ J sin 2x dx 

= — iV cos 8x — J cos 2x + C. 
Illustration 14. 
j cos 7x sin 3x dx = J |[sin (3x + 7x) + sin (3x — 7x)]dx 

= I J sin lOx dx — 5 J sin 4x dx 
= — 2V cos lOx + I cos 4x + C. 



§76] CIRCULAR FUNCTIONS 129 

Illustration 15. 
J cos 4a; cos 7x dx = J | [cos (7x + 4a;) + cos (7x — 4a;)] dx 

= 5 J cos llx dx + hi cos 3a; dx + C 

= -2V sin 11a; + 6 sin 3a; + C. 
Illustration 16. 

J sin 4x sin 2x dx = — ^ J [cos (4a; + 2x) — cos (4a; — 2x)] dx 

— ~ 2] cos 6x dx -{- h j cos 2a; rfx 
= — iV sin 6x + J sin 2x + C. 

Exercises 

In Exercises 1 to 10, verify the differentiation. 

1. y = sin ox, ^ = 5 cos ox, -3-^ = — 2oy. 

dy „ . „ d^y 

2. y = cos 3a;, dx ^ ~ ®'" ' dx^ " ~ ^^* 

3. y = tan 2x, 'd ^ '^ ^^^^ ^^' 

j-^ = 8 sec'' 2x tan 2x. 

dw 

4. 2/ = sin X cos 2x, -r- = cos 2x cos x — 2 sin 2x sm x 

. 3x - 2 dy , 3x - 2 

6. 7/ = sin — ^ — = I cos 



dx ' 5 ' 

dx-^ = - ^' y- 

6. 2/ = tan' 5x, dy = 15 tan^ 5x sec* 5x dx. 

7. y = sec^ 3x, dy = 12 sec* 3x tan 3x dx. 

[y ~ °(1 ~ cos 0), dy = a sin d9. 

Lx = a{d — sin 0), dx = a(l — cos O)d0. 

. /2wt \ dy 2aw /27rt \ 

9. y = a sin I ^^ ^) ' di ^ ^ ^°^ XT' ~ V ' 

10. y = X sin x, dy = (x cos x + sin x) dx. 

11. From the results of Exercise 8, show that -r- = cot s- 

' dx 2 





130 CALCULUS [§76 

Find dy in Exercises 12-20. 

12. y = tan 2xsin 2x. 15. y = cos (3 — x)*. 

sin 2x 



14. y = sin (x^ + 3x — 2). 17. y = x cos 2x — tan 2x. 

18. y = tan- (x — 1). 

19. y = cos* (1 — x» — 2x). 

20. 2/ = sin2 (2x - 1) cos^ (2x - 1). 
Integrate : 

21. dy = sin 2x<fx. 26. dy = sin x cos xdx. 

22. dy — cos 2xdx. 26. dy = tan x sec* xdx. 

23. dy = sec* 4x<ix. 27. dy = Vsin 2x cos 2xdx. 

24. dy = sec 5x tan 5xdx. 28. dy = cos^x sin xdx. 

29. dy = sec* x tan xdx = sec'x sec x tan xdx. 

30. dy = sec" (x — 1) tan (x — 1) dx. 

31. Find the area iinder one arch of the sine curve. 

32. Find the area under one arch of the curve y = 2a* sin* x. 

1 — cos 2x 
Hint, sin* x = „ 

33. The equations of Exercise 8 are the parametric equations of the 
cycloid. Find the length of one arch of the cycloid. 

Hint, ds — -\/{dxy + (dy)". Express ds in terms of e and dO. 

34. Find the area under one arch of the cycloid. 

dv 
36. X = a cos 6, y = a sin 0. Find -j-. Find the length of the 

curve. Find the area bounded by the curve 

36. x = 

the curve. 

dy 

37. X = a cos' <j), y = a sin' </>. Find -j-. Find the length of the 

curve. 

38. Find the volume bounded by the surface obtained by revolv- 
ing y = sin X about the X-axis. 

39. A man walks at the constant rate of 4 feet per second along the 
diameter of a semicircular courtyard whose radius is 50 feet. The 
sun's rays are perpendicular to the diameter. How fast is the man's 
shadow moving along the semicircular wall of the courtyard when he 
is 30 feet from the end of the diameter? 



dv 
36. X = a cos e, y = h sin 6. Find t-. Find the area bounded by 



§77] CIRCULAR FUNCTIONS 131 

40. A drawbridge 25 feet long is raised by chains attached to the end 
of the bridge and passing over a pulley 25 feet above the hinge of the 
bridge. The chain is being drawn in at the rate of 6 feet a minute. 
Horizontal rays of light fall on the bridge and it casts a shadow on a 
vertical wall. How fast is the shadow moving up the wall when 13 
feet of the chain have been drawn in? 

41. Find ^ ii x = y\/y - 1. 

dy .^ 



42. Find ^ ii x = Vl - sin y. 

43. If p' = a* cos 2d . show by implicit differentiation that 
dp _ a* sin 29 

dd^ P 

44. If p 2 cos = a^ sin 36, find ^• 

46. I sin 6x cos 2x dx. 49. I sin Ax cos 7x dx. 

46. I cos 4a; cos 3x dx. 60. I cos 5x cos 9x dx. 

47. I cos 5x sin 2a; dx. 61. I sin wt cos at dl. 

48. I sin Sx sin 3x dx. 62. 1 cos ut cos at dt. 

63. Find the mean ordinate of the curve y = sin x between the 
limits X = and x = tt. 

77. Derivatives of the Inverse Circular Functions. ^ The for- 
mulas for the derivatives of the inverse circular functions are 
readily obtained from those of §§75 and 76. 

. 1 The student will recall that sin"' u is defined for values of u between —1 and +1 
only, and that it is a many valued function. To a given value of u there correspond 
infinitely many angles whose sines are equal to u. This will be seen to be the case on 
sketching the curve y = sin"' u. In this and future discussions of this function it 
will be made single valued by considering only those values of y = sin"' m which lie 

between — ^ and +o'> inclusive. 

The positive sign of the radical in the final formula (1) is chosen because 
cos V ■» \/l — uJ is positive when j/ lies between —^ and +0"* 

Of the functions occurring in (2), (3), (4), (5), and (6), y = cos"' u, and y = sec"' u 
are made single valued by choosing y between and ir, while the remaining functions, 
y — tan"' u, y =• cot"' u, y = esc"' u, are made single valued by choosing y between 

— 2" and +2"' Show that the proper sign has been chosen for the radicals in the 

formulas (2), (5), and (6). 



132 CALCULUS [§77 

Let y = sin"^ u. Then sin y = u. Differentiation gives 

dy du 
cos y -r = T~' 
^ dx dx 

dy _ 1 du 
dx ~ cos y dx 

1 du 

Vl — sin" y dx 

du 
dy dx 



Hence 



dx y'l _ u^ 



Therefore, 



du 



d(sin-iu) dx J/ . , X <*" (1) 

— -— = , » or d(sm-i u) = —7 - '^ ' 

The student will show that 

du 

d(cos-iu) dx J/ , X du (2) 

: = — , > or dtcos-^u) = ,■ ^ ' 

dx . VI - u2 VI - u' 

du 

d(tan-»u) dx ^/* , x <^^ (3) 
or d(tan-'u) = f^p^z 



du 

or d(cot-i u) = - YJ^i (4) 



» or d(sec"^u) = 7 - (5) 

uVu* - 1 



- , ^ or d(csc-i u) = , (6) 

dx uVu" - 1 uVu^ - 1 

Illustration 1. K y = sin"i(x^ — 2x — 3), find dy. By- 
formula (1) 



dx 


" 1 + u^' 




du 


d(cot-» u) 
dx 


dx 

" i + u*' 




du 


d(sec~^ u) 


dx 


dx 


u Vu2 - i 




du 


d(csc-' u) 


dx 



§77] CIRCULAR FUNCTIONS 133 

d(x^ -2x-3) 



dy = 



Vl - {x^ -2x- 3)« 
2(x - l)dx 



Vl - (x* -2x- sy 

Illustration 2. U y = tan~^ 3x, find dy. By formula (3) 

_ djSx) 
^y ~ 1 + (3x)2 
_ 3(Za; 
~ 1 + 9x2" 

Illustration 3. If dt/ = 7-3^ — ^» find y. 



/r 



(ia; 



+ a;2 
or 

y = tan~^ x + C. 

Illustration 4. li dy = ^ , ^ ^ y find r/. 

cix 



■'/r 



+ 9x» 
3(ix 



+ (3x)2 

The expression under the integral sign is now of the form -r-y — i 
whose integral is tan~^ u. Hence 

y = \ tan-i (3x) + C. 

(XX 

Illustration 5. If dy = . . „ g > find ^. 



-/ 



dx 



4 + 9x2 
dx 






+ {ixy 






134 • CALCULUS 

Hence 

y = i tan-i (ix) + C. 

Illustration 6. If dy = — ^ , find « 
V4 - 9x2 »• 



[§77 






9x5 



_ j^ r dx 

~ V Vl-(|x)»' 

= 1.2 r ^dx 

' VVi-(ix) 



The expression under the integral sign is now of the form 

du 
;^/p=| whose integral is sin-» u. Hence 

y = i sin-i (fx) + C. 



Exercises 

dy 
Find ^ in Exercises 1-10. 

1. y = sin-i (x^). 6. 2/ = sin"' (sin x). 

2. y = sin-i (x - 1). f, y = tan-i -^— • 

X — 1 

3. 2/ = tan-Ux2). 8. t/ = sin-» (1 - x)K 

4. y = tan-i (x - 1). 9. y = gec-i (a;2 _ 3). 

5. y = sm (sin-»x). 10. y = a;sin-ix. 
Integrate : 

11. dy = t4%- 

^ 4 + x2 

12. rfw = K-^^- 

^ 9 + x2 

13. rfy = "^^ 



25 + 16x2 



-^ 7 _ dx It 

^ ~ a2 + x^' ^^' y = a *^"~' ^ + ^- 




CIRCULAR FUNCTIONS 135 

dx 



Ans. y = sin~* — f- C 



dx 



20. dy = . — = -• Ans. y = - sec~i - + C. 

Using the results of Exercises 14, 18, and 20 as formulas, evaluate 
the following integrals: 



21. 
22. 
23. 
24. 
26. 
26. 
27. 



J x/16^^^" ^^' J 
J Vl6 - 9x» J ^ 

J Vl - 9x2 J 

J 25TT^ 32. J - 

J 25 + 16x2* 33. J 



dx 



x\/9x2 - 1 
dx 



X2 + 17 

dx 
Vl3 - a;2 
dx 
\/x2 - 19' 
dx 



5x2 + 8 

dx 



J 1 + 16x2" 34- J ^2^43.4.5 J 

J 7v^2^^- 35. J 



^ (x+2)2+l 
dx 



x\/3x2 - 14 



C dx «„ r dx 

28. I -. 36. I , 

J x\/9x2 - 25 J \/9 - x2 

37. Find the area between the ordinates x = 0, x = §, the X-axis 

and the curve y = — . 

Vl -x2 

4a' 

38. Find the area under the curve y = , , . , » above the X-axis, 

" x2 _j. 4o2 ' 



136 



CALCULUS 



[§78 



and between the ordinates x = and x = b. Find the limit of the 
area as b increases without limit. 

39. Find the mean ordinate of the curve y = j--r — ^> between the 

■JT 

limits X = and x = -y 

78. Velocity and Acceleration. If a particle is moving in a 
curved path, its velocity at any point is represented by a vector 

laid off along the tangent with its 
length equal to the magnitude of 

ds 
the velocity, -r:' Thus the ve- 
locity at the point P, Fig. 55, is 
represented by the vector PT. It 
can be resolved into the com- 
ponents PK and PM, parallel to 
the X- and F-axes, respectively. 
~^ These components represent the 
time rates of change of the coor- 
dinates of the moving point P, i.e., 

PK^^"" 




Fig. 55. 



and 



Since 



PM = 



dt 



dy 
dt' 



(1> 



PT = V{PK)^ + {PM)\ 



ds 
dt 



(2) 



This relation can be obtained directly from (2), §63, if we consider 
X and y functions of t. For, we can divide by dt and obtain the 
equation (2). 

In Fig. 55, let PT be the velocity at P, and QT' that at Q. 
Draw from a common origin, o, Fig. 56, the vectors op and oq 
equal to the vectors PT and QT', respectively. Then pq equals 
the vector increment, Av. The average acceleration for the 

Av 
interval At is equal to -rr directed along pq. Lay off, on pq, pm 



§78] 



CIRCULAR FUNCTIONS 



137 



equal to ry As At approaches zero, Q approaches P, and q ap» 

proaches p as indicated by the dotted line, Fig. 56; p7n approaches a 
vector pt directed along the tangent to the arc pq at p. This vector, 

Av 
the limit of — .• represents the acceleration of the particle moving 

in the curved path. Let us calculate its x and y components. In 
Fig. 56, denote: 

op by V and its components by Vx and Vy, 

oq by v' and its components by v'x and v'y, 

pq by Ay and its components by Ay, = v'x — Vx and Ay„ = v'„ — Vy, 

pt by j and its components by jx and j^. 




Fig. 56. 



Then 



Jx = 



lim Ay. _ dv^ 

^'=^0 Af dt 



lim Ay^ 



dv„ 
dt 



dt 
dt 



d^ 
df 



d^ 
dt"^' 



The magnitude and direction of the vector ^ are given by: 



(3) 



(4) 



(5) 



138 



CALCULUS 



[§79 



and 

where is the angle made by pt, Fig, 56, with the positive direction 
of the X-axis. 

Again we can resolve the acceleration j into components along 
the tangent and normal. In Fig. 57, PL is the tangential com- 
ponent and PJ is the normal component. The tangential com- 
ponent clearly produces the change in the magnitude of the 
velocity, and the normal component the change in its direction. 




Fig. 57. 

79. Angular Velocity and Acceleration. If a body is rotating 
about an axis, the amount of rotation is given by the angle d 
through which a line in the body turns which intersects the axis 
and is perpendicular to it. Thus in the case of a wheel the rotation 
is measured by the angle 6 through which a spoke turns. ^ is a 
function of the time t. The rotation is uniform if the body rotates 
through equal angles in equal intervals of time. If the uniform 
rate of rotation is co radians per second, the body rotates through 
6 = cat radians in t seconds. If the rotation is not uniform the 
rate at which the body is rotating at any instant, the angular 
velocity, is 

<^^M^o^t dt' 



§791 CIRCULAR FUNCTIONS 139 

Similarly, the angular acceleration a is the time rate of change of 
the angular velocity. Then, 

do) dW 
"~lt ~ dt^' 

If we consider a particle at a distance r from the axis of rotation, 
its linear velocity v is 

V = cor 

and is directed along the tangent to the circle described by the 
particle. The tangential acceleration is 

jt = ar. 

Exercises 

, 1. The following formulas have been established for linear motion, 
with constant acceleration : 

V = Vo + jt. 
S = Vot + ^jl\ 

I' - ^* = js. (See §38.) 

Show that the corresponding formulas for rotation are: 

e = cjoi + W^' 

W^ COo^ 

T - "2- = «^- 

2. A flywheel 10 feet in diameter makes 25 revolutions a minute. 
What is the linear velocity of a point on the rim? 

3. Find the constant acceleration, such as the retardation caused 
by a brake, which would bring this wheel to rest in 30 seconds. How 
many revolutions would it make before coming to rest? 

4. A resistance retards the motion of a wheel at the rate of 0.5 
radian per second per second. If the wheel is running at the rate of 10 
revolutions a second when the resistance begins to act, how many 
revolutions will it make before stopping? 

5. A wheel of radius r is rotating with the uniform angular velocity 
w. Find the direction and magnitude of the acceleration of a point 
on the rim. 



140 



CALCULUS 



[§80 



Hint. The coordinates of the point can be written x = r cos w<, 
y = rain. uL Find -jtj and tt^* 

6. A wheel of radius r is rolling with the uniform angular velocity 
u along a horizontal surface without slipping. How fast is the axle 
moving forward? The parametric equations of a point P on the rim 
are: 

X = r{u}t — sin cat) 
y = r(l — cos (at). 

Find the magnitude and the direction of the velocity of P at any 
instant. What is the velocity of a point at the top of the wheel? 
At the bottom? 

7. If a particle moves in such a way that its coordinates are 
re = a cos t -\- h, y = a sin < + c, where t denotes time, find the 

equation of the path and show that the par- 
j> tide moves with constant tangential velocity. 

80. Simple Harmonic Motion. Let the 

point P, Fig. 58, move upon the circumfer- 
ence of a circle of radius a feet with the 
uniform velocity of v feet per second, so that 

V 

the radius OP rotates at the rate of - = co 
Fig. 58. o, 

radians per second. The projection, B, of P 
on the vertical diameter moves up and down. If the point P 
was at C when t = 0, the displacement, OB = y, is given by 

y = asind = asinwf. 

If the point P was at D when t = 0, we have 

y = asin(co^ — a). (1) 

Any motion such that the displacement at time t is given by (1) 
is called a simple harmonic motion. Thus the point B, Fig. 58, 
describes simple harmonic motion. The abbreviation "S.H.M." 
will be used for "simple harmonic motion." 

From (1) it follows that the velocity of a point describing 

S.H.M. is 

dy 
dt 




(2) 



§81] CIRCULAR FUNCTIONS 141 

and that the acceleration is 

-77^ = —a cousin (co^ — a). (3) 



The second member is —oihj, by equation (1), Hence 

- a)2y, (4) 



or 

df" 



+ co^y = 0. (5) 



Equation (4) shows that the acceleration of a particle describing 
S.H.M. is proportional to the displacement and oppositely di- 
rected. That the acceleration is oppositely directed to the 
displacement is to be expected from the character of the motion, 
which is an oscillation about a position of equilibrium. Thus if 
the body is above this position the force is directed downward, 
and vice versa. In Fig. 58, the point B has a positive acceleration 
when below and a negative acceleration when above 0. The 
acceleration is zero at 0, a maximum at the lower end of the diame- 
ter, and a minimum at the upper end. 

In accordance with (2) the velocity is zero at the two ends of the 
diameter. The velocity has its greatest numerical value when 
B passes through in either direction. 

Equation (4), or (5), is called the differential equation of 
S.H.M. The proportionality factor co^ is connected with the 

period T by the relation T = — The equation (4) was 

obtained from (1). Frequently it is desired to solve the converse 
problem, viz., to find the motion of a particle whose acceleration is 
proportional to the displacement and oppositely directed. In 
other words, a relation between y and t is sought which satisfies 
equation (4). Clearly (1) is such a relation. However, it will be 
instructive to obtain this relation directly from (4). 

First, a differential equation equivalent to (4) will be obtained 
in the solution of the problem of the motion of the simple pendulum. 

81. The Simple Pendulum. Let P, Fig. 59, be a position of the 
bob of a simple pendulum at a given instant and let it be moving 
to the right. If s denotes the displacement considered positive 



142 



CALCULUS 



[§81 



on the right of the position of equiUbrium, -tt^ is the acceleration 

ds 
in the direction of the tangent PT, for ^ is the velocity along the 

tangent. This acceleration must be equal 
to the tangential component of the accelera- 
tion due to gravity, if the resistance of the 
air be neglected. This component is equal to 
— g sin 9. Since it acts in a direction opposite 
to that in which s is increasing, it must be 
taken mth the negative sign, i.e., the acceler- 
ation diminishes the velocity. We have then 




Fia. 59. 



d's . . 



(1) 



If the angle through which the pendulum swings is small, 
sin 6 can be replaced by 6. Then (1) becomes 

d^ 
dt^ 
Since s = W, 



(2) 



d^ 
dV 



I 



(3) 



Putting , = w^ for convenience in writing, 
d-'d 



dt^ 



= - co^d. 



dd 



Multiplying by 2 -ir and integrating. 



\dt) - 



oiW^ + C». 



(4) 



The arbitrary constant is written for convenience in the form C. 
The constant must be positive. Otherwise the velocity would be 
imaginary. Extracting the square root, 
dd 



dt 



= VC*- w2 0»' 



or 



dd 



Vc^ - w' d^ 



= dt. 



(5) 



§81] CIRCULAR FUNCTIONS 143 

Integration gives 

— sin~^ -p7 = t -\- Ci. 

sin~^ -^ = cof + coCi 

= «< + C2, 
where the constant coCi is replaced by the constant C2. Then 
-T7 = sin (wf + C2), 

= — sin (a>« + C2) 

CO 

= C3 sin {wt + C2), 
where — has been replaced by C3. Therefore 

CO 

e = Ci sin (co< + C2) (6) 

is the equation of the angular displacement of the pendulum. 

2_. /r 

The form of (6) shows that the motion is of period — = 2ir'\^-' 

It is a S.H.M. and contains two arbitrary constants. They can 
be determined by two conditions, e.g., the displacement and 
velocity at a given instant. Suppose the bob drawn aside to the 
right so that the string makes an angle 9o with the vertical. The 
bob is then released without being given an impulse; i.e., with an 
initial velocity zero. The time will be counted from the instant 
of release. The conditions are then 

d = do (7) 

and 

f = » («) 

when t = 6. From (6), 

do 

-jj = C0C3 cos (cof + Ci). 

The condition (8) gives 

= C0C3 cos C2, 

or cos C2 = 0. Whence C2 = ^- Then (6) becomes 
6 = Cz cos cot. 



144 CALCULUS [§81 

The condition (7) gives 

do = Cz. 
Hence 

e = do cos oit. (9) 

Multiplying by I and recalling that W = s, and denoting Ido by 
So, we have as the equation for the displacement s, 

S = So cos bit. (10) 

The period is T = — = 27r 'y-. When is the velocity of the 

bob greatest? When least, numerically? 

Equation (6), the solution of (4), shows that, if the acceleration 
of a particle is proportional to its displacement and oppositely 
directed, the particle describes S.H.M. 

Exercises 

1. Write the differential equations of the following simple harmonio 
motions. Find the period in each case. 

y = 5 sin 3^ 

2/ = 6 sin Izt + Ij • 

y = 5 cos 3/. 

2/ = 4 sin 2t + S cos 2t. 
J/ = 7 sin {8t + a). 

2. Write the equation of a S.H.M. which satisfies the equations: 

J + 3!,-0. 



CHAPTER IX 

EXPONENTIAL AND LOGARITHMIC FUNCTIONS 

82. Derivative of the Exponential and Logarithmic Functions. 
Let 

T/ = a*. (1) 

Then 

Ay = a'(a^ - 1) 
Ay _ ^/(i^ — 1\ 
Ax~ \ Ax ) 

dx~ ^ Ax^o Aa; ^'^> 

„, a^ — I . . . . , lim fl"^ — 1 . 

bince — -r IS independent of x, ^^q — r is a constant 

for a given value of a. Call this constant K, so that 
Then from (2), 



_ lim «^- 1 ,^. 



% = Ka: (4, 

Equation (4) shows that the slope of the curve y = a'' is propor- 
tional to the ordinate of the curve. In other words, the rate of 
increase of the exponential function is proportional to the function 
itself. 

When a; = 0, it follows from (2) and (3) that 

dy Hm a^ - 1 



dx ^-0 Ax 



= K. 



Consequently the constant K introduced above is the slope of the 
curve t/ = a^ at the point (0, 1). This slope depends upon the 
value of a. Let e be that particular value of a for which the 
corresponding curve, y = e^, has a slope equal to 1 at the point 
where it crosses the F-axis. 

10 145 



146 CALCULUS [§82 

If, then, 



equation (4) becomes 



y = 6"=, (6) 

dy 



dx 



since K = \'\n this case. Or 

de 



Then 
and 



dx =''• (6) 

de» du _. 



de» = cdu. (8) 



Equation (6) shows that the slope of the curve y = e' is equal 
to the ordinate of the curve. The number e is the base of the 
natural system of logarithms. It is sometimes called the Naperian 
base. Its value, 2.71828 • • • • , will be calculated later in 
the course. 

The formula for the derivative of the natural logarithm of a 
function is now easily obtained. In calculus if no base is indi- 
cated, the natural base is understood. Thus log u means log, u. 

If 

y =- log u, 
u = e« 



du dy 



and by (7) 
Whence 



That is 

or (9) 



dx '' 


= ^"d|- 


dy 
dx ' 


_ 1 du 
~ e" dx 


= 


_ 1 du 
~ u dx 


d(log 


u) _ 1 du 


dx 


~ u dx' 


d(Iog 1 


du 

u) = — • 
' u 



§82] EXPONENTIAL AND LOGARITHMIC FUNCTIONS 147 

Since ^ 

logoW = logoclogw, (10) 

d (logau) , „ ^ 1 du 

or (11) 

du 

d(logaU) = logaC — ' 

11 y = a", log y = uloga 

l dy _ , du 
y dx ~ dx 

dy . du 

Tx^^y^^'^^'Tx 

. du 
= aMoga^. 



That is 



da" , du 

^ = auloga^ 

da« = a«» log a du 



(12) 



Illustrations. 

1. If 2/ = e*', dy = e^'d(x2) = 2xe''"dx. 

2. If 2/ = e''"" ", dy = e "''^ ''d{smx) = cos x e"° * dx. 

r. Tr 1 / . HX 1 1 ^(^ + 1) 1 dx 

3.Uy = logio (x + 1), dy = logio e ^ , ^ = logio e^-^y 



•l. J.1 J/ = lUJ 


U^ t 


- ^), 




t 


''V 




x + V 


»Let 










z 


= 


log u 


Then 










& 


= 


u 


Taking logarithms 


to the 


base 


1, 
















z 


log 


a e 


= 


logo U 


That is 






















log 


u 1 


og» 


1 e 


= logo M 



148 CALCULUS [§82 

5. Uy = e*^'^"'^ dy = e*'*'^'"^ d(tan-i x) = e*^"" '^ rr^ • 

1 -\-x 

6.1iy = log 1^^, y = 2 log (1 + x) - 3 log (1 - X), 

and 

, 2dT; , Sdx 5 + X , 

7. If 2/ = e* sin x, -r- = e*(cos x + sin x) 



and 



-j-^ = 2e* cos X. 



Exercises 



Find the first derivative of each of the following : 

1. y = e'*. 6. 2/ = log (1 - x^). 11. y = e^''^*. 

2. y = e'\ 7. y = e' cos x. 12. y = e***^ ^'. 

Z. y = log (x"). 8. 2/ = e'». 13. y = log Vx'' - 1. 

4. y = log (x'). 9. y = e'=^', 14. y = e~* sin x. 

6. y = log (x* - 1). 10. y = log [^^^I' 16- 2/ = 10'- 

16. y = logio X. 

17. y = 5'. 

18. y = x^ 5*. 

19. Show that the aubtangent for the curve y = a'' is a constant. 
What is this length when a = e? 

Illustrations. 

8. If dy = e'rfx, y = \ e'dx = e' -\- C. 

9. If dy = xe'^dx. 

y = \ xe'^dx 

= |e'' + C. 
10. If dy = ^> 

X 

J/ = log X + C 
= log X + log i2! 
= log Kx. 



§82] EXPONENTIAL AND LOGARITHMIC FUNCTIONS 149 

dx 

11. If dy = ^:^, 

y = \og{x + l) + C 
= log K(x + 1). 

xdx 

12. If dy = ^^^p3, 






a;2 + l 
2xdx 



x' + l 
= i log (x^ + 1) + (7 
= log Vx2 + 1 + C 
= log KVx^ + 1. 

13. If dy = e™* cos xdx, 

y = j e "° * COS X dx 

1.1 Tf J (x + l)dx 

14. If dy = ^2 + 2x + 3' 



-/ 



(x + l)dx 



x2 + 2x + 3 

r 2(x + l)dx 
■^ ^ J x2 + 2x + 3 

= h log (x2 + 2x + 3) + (7 

= log Vx^ + 2x + 3 + C 

= log KVx^ + 2x + 3. 

15. If dy = tan x dx, 

y = I tan x dx 



-/- 



sin X , 

ax 

cosx 



= — log cos X + (7 
= log sec X + C. 



150 



CALCULUS 



l§82 



16. If 



17. If 



18. Find 
Let 

whence 
and 



Then, since 



dy = cot xdx, 
y = \ cot X dx 



Si 



dx 



sin X 
= log sin X -^ C. 
dy = sec x dx, 
y = J secx dx -\- C 

(sec x + tan x) sec xdx 
sec X + tan x 

/-■'■■""•" 



f- 



dx 



sec X -J- tan x 
= log (sec X + tan x) + C 
dx 



S 



Vx^ ± a2 



v^ = x^ ± a" 
2v dv = 2x dx 



dv 

X 

dv 

X 



dx 

dx 

V 



dx _ Ci 



dx + dv 
X + V ' 

dx + c^p 



19. If 



Vx^ ± a2 J X + v 

= log (x + v) -\-C 

= log (a; + Vx* ± a2) + C. 



y 



■/. 



e' + 1 
= log (e* + 1) + C. 



EXPONENTIAL AND LOGARITHMIC FUNCTIONS 151 

20. If ^^ = ^, 

y X 

log 2/ = log X + log C 
log y = log Cx 
y = Cx. 

21. If — = n — J 

y X 

log y = n log a; + log C 
= log a;» + log C 
= log Cx", 
1/ = Cx". 

Exercises 

The results of Illustrations 15, 16, 17, and 18 are to be used as 
formulas of integration. 

In the following exercises and y : 

20. dy = x2 e*' dx. 27. dy = tan 2x dx. 

21. dy = e*-"" sec"^ X dx. 28. dy = cot2xrfx. 

22. dy = — r-r. 29. dy = sec 2x dx. 

" X + 1 " 

23. dy = Y^x ^°- "^^ = (logx)*^. 

_, - xdx «^ , cos X dx 

2*- ^^ = IT^^' 31. dy = ^-j^^^^^- 

26. d2/ = ^,- 32. d2/ = (!l-ZiZM£. 

'^ 1 — x^ -^ e* + e * 

«/. , (e"^ + l)dx „„ , sec^xdx 

(e2* - l)dx 



34. dy = 



+ 1 



35. Find the area between the equilateral hyperbola xy = 10, the 
X-axis, and the lines x = 1 and x = 2. 

36. The volume of a gas in a cylinder of cross section A expands 
from volume vi to volume Vi. If it expands without change in tem- 
perature the pressure, p, on the piston varies inversely as the volume 
(Boyle's Law, pv = K). Show that the work done by the expansion is 



pdv 



K\ ? = xiog^;. 



152 CALCULUS [§82 

37. The subtangent of a curve is of constant length, k. Find the 
equation of the curve. 

38. For what value of x is the rate of change of logio x the same 
as the rate of change of z? 



-I 

I tan 36 dd. 48. | 



39. I CSC X dx. (See Illustration 17.) 

xdx 



41. j e*"'^'' sin e dd. 49. j lO'dx. 

42. I cot I dd. 60. I 



cos^ (5x — 4) 
2 (1 — Bmx)dx 



43. j X sec2(z2 + l)dx. 51. | 
I ' I a; + cos X 

44. I sec I dx. 52. j e^^ + """^ "> sin dd. 

Cdx C xHx 

46. I — 63. I ; , 3 - 

46. f'-^^. 64. f-^:^. 
J„ Vl+x^ J Va^ + X* 

66. I (tan 20 — \)^de = \ tan 20 + log cos 20 + C. 



/ 

/Idx 



57. I C8c(7x + 5) dx. 

68. 1 -^ +^)^" 

Vx^ + 4x + 7 

69. 

~ (3x + 2)dx 
"• ' 3x* + 4x + 9 



§83] EXPONENTIAL AND LOGARITHMIC FUNCTIONS 153 

(x + 2)dx 



61 



/. 



(x" + 4x + 7)2 

62. Show that y = ae~** cos o)t satisfies the differential equation 

g+2fc* + („.+*=)„-0. 

63. Find the mean ordinate of the curve w = - between the limits 

" X 

X = 1 and X = 2. 

83. Logarithinic Differentiation. It is often advantageous in 
finding the derivative oi y = f{x) to take the logarithm of each 
member before differentiating. A number of examples will be 
solved to illustrate the process. 



(x- 1)3 



lUllSLIUUUII, 1. 


X' lllU 


Liie ut 


{x + 1)^ 






y = 


{x - 1)^ 
(x+ 1)' 


and take the logarithm 


of each member. 


log 3 


!/ = 1 log (X 


- 1) - 1 log (x + 1). 


Differentiating, 


Idy 
y dx 




2 3 




3(x 


- 1) 5(x + l) 



Let 



x + 19 



15 (a;2 - 1) 
dy _ g + 19 
dx ~ 15 (a;2 - 1) 



2/ 



X + 19 (x - 1)^ 





15 (x2- 1) (x + 1)' 
x + 19 




15 (x - l)^ (x + 1)^ 


1 11.11 RtTniinti. 2 


•\/l — x^ 
Find the derivative of ■" 




i/x^ + i 




Vl-a;^ 

2/ - 3/-^^^^ 



154 CALCULUS l§83 

log 2/ = § log (1 — x2) — A log (x2 4- 1) 
1 dy X 2x 

ydx^~ 1 -x2 ~ 3 (a;2 + 1) 

_ _ 3^ (5 + a:") 
~ 3 (1 - x^) 
dy a; (5 + a:') Vl - x^ 

dx ~ 3(1 -x^) v^mh: 

a: (5 + a;2) 



3Vl -a;2(x2+ 1)^ 

This method is manifestly shorter and simpler than that of differ- 
entiating by the rule for the derivative of a quotient. 
Illustration 3. Find the derivative of {x^ + l)''''^^. 
y = {x"^ -\- 1)3^^2 
log y = (3a; + 2) log {x^ + 1) 



dy 



J^=(^^ + 2)^-^ + 3log(x^ + l) 



^^ = [(3a: + 2) ^^ + 3 log (x2 + 1)] {x^ + 1)3-+^ 

- -p is called the logarithmic derivative of y with respect to x. 
It will be considered further in a later article. 

Exercises 
Find the derivative in Exercises 1-8. 

3 

1. 2/ = ^^^tilL- 3. 2/ = (X + 1)3 (2x + 5)1 

(x-7)5 

2- ^ = (x-4)t(r5)^ - *• ^ = ^(1 + ^)^^^- 

5. 2/ = x"n^. (Solve by two methods.) 

6. 7/ = x^^°^. 

7. s = (7< + 3)10^-2. 

8. 2/ = xV'^. 

In Exercises 9-16 find the logarithmic derivative. 

9. 2/ = e^'. 12. y = x". 
10. y = x\ 13. y = ex". 



§84] EXPONENTIAL AND LOGARITHMIC FUNCTIONS 155 

11. y = Zi' 14. 2/ = e*'+^ = ce**. 

15. y = 10*'+^. 

16. y = uv, where u and v are functions of x. 

17. y = uvw, where u, v, and w are functions of x. Find t— 
Find y if its logarithmic derivative is: 



22. A;. 

(1) 

(2) 

Equation (2) expresses the fact already noted in §82, as a 
characteristic property of the exponential function, viz., that the 
function increases at a rate proportional to itself. We can show, 
conversely, that if a function increases at a rate proportional to 
itself, it is an exponential function. 

Thus, let it be given that 
dy 



18. 61 + 7. 


Ans. y = Ce ' 


X 




19 J. 

X 




20. - 

X 


21. 


FkxY 


84. 


Compound Interest Law. If 








y = 


Ce*S 








dy 
dt 


CA;e*' 


= ky. 



dt = ^y- (3> 



Then 

dy 



kdt 



logy = kt + C 

2/ = e*'+c = e^e*'. 
Hence 

y = Cie*'. (4) 

When a function varies according to this law it is said to follow 
the "compound interest law." For, if a sum of money be placed 
at compound interest, its rate of increase, for any interest term, is 
proportional to the amount accumulated at the beginning of that 
term. The more frequently the interest is compounded the 
more nearly does the amount accumulated increase according 
to the exponential law. 

In many cases in nature the function decreases at a rate pro- 



156 CALCULUS 

portional to itself. The compound interest law appears in this 
case in the form Ce~*', where A is a positive constant. For, if 

^= -kv 

it follows that 

y = Ce-*«. 

Illustration 1. Newton's law of cooling states that the tem- 
perature of a heated body surrounded by a medium of constant 
temperature decreases at a rate proportional to the difference in 
temperature between the body and the medium. Let d denote the 
difference in temperature. Then 

ft = - ^^- (S) 

and 

e = Ce-*«. (6) 

The meaning of the constant C is seen at once on setting t = 0. 
It is the difference in temperature between the body and the 
medium at the time t = 0. If this initial difference in temperature 
is known, (6) gives the temperature of the body at any later 
instant. Call the initial difference in temperature ^o- (6) 
becomes 

e = doe-"'. (7) 

The time which is required for the difference in temperature to 
fall from di to 62 can be found from (7). Thus 
01 = 0oe-*'i 
62 = ^oe-*'» 






whence 



/.-<. = ^ log |. (8) 

This result could have been obtained directly from the differ- 
ential equation (5). Thus 

..= -if. (9) 



§84] EXPONENTIAL AND LOGARITHMIC FUNCTIONS 157 



Integrating the left-hand member between the limits ti and <2 and 
the right-hand member between the limits di and 62, 



f— iX 



e 



Illustration 2. Find the law of variation of the atmospheric 
pressure with height. 

Consider a column of air of unit 
cross section (Fig. 60). Denote 
height above sea level by h and the 
pressure on unit cross section at this 
height by p. The difference in pres- 
sure at C and D is the weight of the 
gas within the element of volume 
of height Ah. 
Thus 

Ap = — gp Ah, 

where p is the average density of the air in the volume CDEF. 
Then 






Fig. 60. 



and 



Ap 

Ah 

dp 
dh 



= - gp 



= - gp, 



(1) 



where p is the density at C. If the temperature is assumed con- 
stant, the air obeys Boyle's Law, pv = c, where v denotes the 
volume occupied bj'^ unit mass of air. Since 

mass _ 1^ _ P 
volume V c 



P = 



dh 



= - kp, 



158 


CALCULUS 


where 






* = ^ 


Integration gives 





[§84 



logp = — kh + logCi, 
or 

When h = 0, p = Po, the pressure at sea level, and Ci = po- 

Hence 

p = poe~**. (2) 

If h is measured in meters and p in millimeters of mercury, 
k = risVir, (2) becomes 

p = 760e-iToT- (3) 

Exercises 

1. A law for the velocity of chemical reactions states that the 
amount of chemical change per unit of time is proportional to the 
mass of changing substance present in the system. The rate at which 
the change takes place is proportional to the mass of the substance 
still unchanged. If q denotes the original mass, find an expression 
for the mass remaining unchanged after a time t has elapsed. 

2. Assuming that the retardation of a boat moving in still water is 
proportional to the velocity, find the distance passed over in time t 
after the engine was shut off, if the boat was moving at the rate of 7 

7 
miles per hour at that time. Ans. s = t(1 — e *'). 

3. The number of bacteria per cubic centimeter of culture increases 
under proper conditions at a rate proportional to the number present. 
Find an expression for the number present at the end of time t. Find 
the time required for the number per cubic centimeter to increase 
from 6i to 62- Does this time depend on the number present at the 
time t = 0? 

4. A disk is rotating about a vertical axis in a liquid. If the retar- 
dation due to friction of the liquid is proportional to the angular 
velocity «, find w after t seconds if the initial angular velocity was wo. 

6. If the disk of Exercise 4 is rotating very rapidly, the retardation 
is proportional to w^ Find w after t seconds if the initial angular 
velocity was wq. 



§85] EXPONENTIAL AND LOGARITHMIC FUNCTIONS 159 

85. Relative Rate of Increase. If the rate of change of a func- 
tion is divided by the function itself, the quotient is the rate of 
change of the function per unit value of the function. This 
quotient has been called the relative rate of increase of the function. 
If a function varies according to the compound interest law, its 
relative rate of increase is constant, i.e., 

y dt 

One hundred times the relative rate of increase is the percent rate 
of increase. Thus if 

- t = 0.02, 

V dt ' 

the percent rate of increase is 2. This means that y increases 2 
percent per unit time. Any of the Exercises 1-5 might have been 
stated in terms of the relative rate of increase of the function 
concerned. 

Exercises 

1. Given that the intensity of light is diminished 2 percent by 
passing through one millimeter of glass, find the intensity / as a func- 
tion of t, the thickness of the glass through which the light passes. 

2. The temperature of a body cooling according to Newton's Law 
fell from 30° to 18° in 6 minutes. Find the percent rate of decrease of 
temperature per minute. 

86. Hyperbolic Functions. The engineering student is likely 
to meet in his reading certain functions called the hyperbolic 
functions. These functions present analogies with the circular 
functions and they are called hyperbolic sine, written sinh, hyper- 
bolic cosine, written cosh, and so on. 

These functions are defined by the equations: 

cosh X = — n — ' sech x = 

. , e' — e~' , 

smh X = — -X » csch x = 



2 

gx — g-x 

6' — e~* 



tanh X = .. , ,_. ' coth x = 
odd functions 




e* 4- e~* tanh x 

cosh X and sech x are even functions, while the remaining four are 



160 CALCULUS [§87 

Exercises 

1. By making use of the definitions the student will show that the 
following identities hold. They are analogous to those satisfied by the 
circular functions. 

cosh^x — sinh^x = 1. 
1 — tanh*x = sech^x. 

2. Show by the use of the defining equations that : 

d cosh X 



dx 
d sinh X 

dx 
d tanh x 

dx 
d coth X 

dx 
d sech X 

dx 
d csch X 



sinh X, 
= cosh X. 
= sech^ X. 
= — csch^ X. 
= — sech X tanh x. 
= — csch X coth X. 



dx 

3. Sketch the curves y = cosh x, y = sinh x, and y = tanh x. 

87. Inverse Hyperbolic Functions. The logarithms of certain 
functions can be expressed in terms of inverse hyperbolic functions 

Let 

y = sinh~* x. 

e« — e~y 





J/ — Buiii (/ — 2 


or 






e'^y — 2xey — 1=0, 


whence 






c = X ± Va;^ + 1. 



The minus sign cannot be taken since e" is always positive. 
Hence 

e" = a; + Vx^ + 1 , 
and 

y = sinh-' x = log (x + Va;^ + !)• 



1. Show that 
Since 



§88] EXPONENTIAL AND LOGARITHMIC FUNCTIONS 161 

Exercises 

cosh~i a; = log (x + va;" — 1). 

X - -y/x^ - 1 = 7 > 

X + Vx'^ - 1 

log (x - a/x" - 1) = - log {x + \/x2 - 1). 
Therefore 

cosh~ia; = + log (x + 's/x^ — 1). 

The inverse hyperbolic cosine is then not single valued. Two values 

of cosh~i x, equal numerically but of opposite sign, correspond to each 

value of x greater than 1. 

2. Show that: 

1 + x 
tanh'^x = § log , _ > if x'^ < 1; 

X + 1 
coth~i X = 2 log _ ^ » if x^ > 1 ; 

1 -f. a/1 _ a;2 

sech~* X = ± log > if < x ^ 1 ; 



and 



, _. , 1 + Vx^ + 1 .. 

csch ^x = log > if X > 0: 

^ X ' 



csch ^ X = log > if X < 0. 



The student is not advised to memorize the formulas of this and the 
preceding sections at this stage in his course, but to acquire sufficient 
familiarity with the hyperbolic functions to enable him to operate with 
these functions by referring to the definitions and formulas given here 
and to others that he will find in mathematical tables. 

88. The Catenary. Let AOB, Fig. 61 a, be a cable suspended 
from the points A and B and carrying only its own weight. Let 
us find the equation of the curve assumed by the cable, consider- 
ing it homogeneous. We shall assume that the curve has a 
vertical line of symmetry, OY, and that the tangent line drawn 
to the curve at is horizontal. 

Take 07 as the F-axis. Imagine a portion of the curve, OP, 
of length s, cut free. To hold this portion in equilibrium the 
forces // and T, Fig. 61 h, must be introduced at the cut ends. 
n 



162 



CALCULUS 



H and T are, respectively, equal to the tension in the cable at 
the points and P and they act in the direction of the tangent 
lines drawn to the curve at these points. The portion of the cable 
OP, Fig. 61 b, is in equilibrium. Hence H', the horizontal com- 
ponent of T, is equal to H. 




V, the vertical component of T, must balance the weight of the 
portion OP of the cable. Hence 

V = sw, 
where w is the weight of a unit length of the cable. 
From Fig. 61 6, it is seen that 

dy _ V^ _ V^ _ ws 
dx~ H' ~ H ~ H' 
Let 

w _ 1 
H ~ a 
Then 

dy s 

di " a' (^) 

This differential equation involves three variables, viz., x, y, and 
8. s may be eliminated by differentiating and substituting for ^ 
its value, 



Thus 



dx 



a dx a \^ + W 



EXPONENTIAL AND LOGARITHMIC FUNCTIONS 163 
The equation now involves only two variables and may be written 



, dy 

ok upon -j- 

tion (2) is 



= -dx. (2) 



dv 
If we look upon -r- as the variable u, the left-hand side of equa- 



du 



vTTw2 

whose integral is log (u + \/l -|- u^). (See Illustration 18, §82.) 
Integrating (2), 

When a; = 0, ^ = 0. 

Hence C = and (3) becomes 



dv 
From the symmetry of the curve -r- changes sign when x is re- 
placed by —X. Then from (4), 

Subtracting (5) from (4), 

2g=e«-e"a, (6) 

or 

^ = sinh? (7) 

dx a 

Integrating (7), 

y = a cosh - -f- d- (8) 

If the origin is taken a units below the point 0, Fig. 61 a, 
y = a when x = 0, and C2 = 0. Hence 

y = a cosh — (9) 



164 CALCULUS 

This is the equation of the curve assumed by the cable. It 
is called the catenary. 

Equation (9) can be written 

Y = cosh X, (10) 

where 

r = ^ and X = ?• 
a a 

The constant a depends upon the tautness of the cable. Equa- 
tion (10) shows that the curve y = cosh z if magnified the proper 
number of diameters will fit any cable hanging under its own 
weight. 

The length of OP can be found by substituting in formula 2, §63, 

du 
the value of j- given by (7), and integrating. 



-^' 



ds = « / 1 + sinh' - dx 
a 



1C 

= cosh- dx. 
a 

X 

s = a sinh + C3. 
a 

Since s is measured from the point where the curve crosses the 
y-axis, s = when x = 0. Hence Cj = and 

s= a sinh -• (11) 

a 

Exercises 

1. If the two supports A and B, Fig. 61, are on a level, L feet apart, 
and if the sag is d feet, show that the tension, T, in the cable at the 
points A and B is 

T = w(a + d). 

2. Beginning with equation (6) find expressions for y and s without 
making use of hyperbolic functions. 

3. If the cable. Fig. 61, is drawn very taut, show that the equation 
of its curve is approximately 

X2 



y = 



2a 



if the origin of coordinates is taken at the lowest point of the cable. 

Hint. Begin with equation (2) and note that i-r-j is small com- 
pared with 1. 



CHAPTER X 
MAXIMA AND MINIMA 

In previous chapters maximum and minimum values of func- 
tions have been found by making use of the derivative. Besides 
this method several others which do not involve the use of the 
derivative may, at times, be used to advantage. 

89. The Maximum or Minimum of y = ax^ + j8x + 7. In 
elementary analysis the student learned that y = az^ + j8a; + 7 
represents a parabola with its axis parallel to the F-axis, and 
that the equation can be put in the form y = a(x — p)^ -\- q. 
The point (p, q) is the vertex of the parabola. If a is positive, 
the vertex is a minimum point, if negative, a maximum point of 
the curve. Let 

y = Sx^ - 12x + 19. 
y = 3ix- 2)2 H- 7. 

The last equation shows at once that the minimum value of the 
function is 7 and that it occurs when x = 2. 

Exercises 

Find the maximum or minimum values of the following: 

1. y = 3x2 _ 2x + 1. 

2. y = 3x - 2x2 + 1. 

3. y = 3x2 _^ 7a., 

4. If a body is thrown vertically upward with an initial velocity of 
a feet per second, its height h in feet at the end of t seconds is given by 

h = at - 16.1<2. 

To what height will the body rise if thrown with an initial velocity of 
32.2 feet per second? When will it reach this height? 

90. The Function a cos x + b sin x. The function a cos x + 
6 sin X is of frequent occurrence. If it is put in the form of 

165 



166 



CALCULUS 



^90 



the product of a constant by the cosine of a variable angle, the 
maximum and minimum values can be found at once. Thus 



cos X + —r= 



a cos X -\-hsmx = ^aP' + h''' 

a b 

Now, — / and — , 

'Va^ + b^ Va^ + b^ 

sine, respectively, of an angle a. For if P, Fig. 
62, be the point (a, 6) and the angle POX be a, 



Va^ + 62 



sin X 



[Va^ + b^ 

may be regarded as the cosine and 



and 



Va^ + b^* 
b 




Va^ + b^ Fig. 62. 

Hence 

a cos X -\- b sin x = y/a^ -\- b^ (cos x cos a + sin x sin a), 



or 



a cos X -r b sinx = -x/a^ -j- b^ cos (x — a). 



(1) 



The quadrant in which ce lies will be determined by the signs of 
a and b. 

a is in the first quadrant if a is positive and 6 is positive. 
a is in the second quadrant if a is negative and b is positive. 
a is in the third quadrant if a is negative and b is negative. 
a is in the fourth quadrant if a is positive and b is negative. 

In polar coordinates equation (1) shows that the function 
a cos X + bsinx is represented by a circle passing through the 
pole, of diameter -y/a^ + b^, and with its center on the line making 
an angle a with the polar axis. 

The right-hand side of equation (1) shows that the function is 
represented graphically in rectangular coordinates by a cosine 
curve of amplitude -y/a^ -f b^. Thus, the maximum value of 
a cos X + & sin a; is \/a^ + b^ and occurs when x = a. The mini- 
mum value of the function is — \/a'^ + b^ and occurs when 
2 = a -|- X. 

Two examples giving rise to this function are solved below. 



§90] MAXIMA AND MINIMA 167 

Illustration 1. The weight W, Fig. 63, rests upon the horizontal 
surface AB. P is the force, inclined at an angle 6 with the 
horizontal, which will just cause the weight to slide over the 
plane. The problem is to find the angle 6 for which P will be a 
minimum. The coefficient of friction is denoted by ^i. 

The normal pressure, N, between the weight and the plane is 
(]F — P sin 0), the difference between W and the vertical com- 
ponent of P. The force of friction, F, is then fi {W — P sind). 
The horizontal component of P equals F. Hence 

niW - Psind) = Pcosd, 
or 

^ = cos0+Atsine. (2) 

uW 
Since ix and W are constants,-p- is a maximum when and only 

when P is a minimum. Hence to find the minimum value of P 

uW 
we may find the maximum value of -p- and multiply its reciprocal 

hy fiW 



W 




p 

B 



A < — i:i^ i tiw^m i B 




N 

Fig. 63. Fig. 64. 



From (2), 



^ = Vl + M^' cos (0 - a), 



where a is an angle in the first quadrant whose tangent is /x, 
the coefficient of friction. Therefore, when 6 is acute and equal 

to tan~^ /x, JP is a minimum and equal to , 

Vl + /X'' 

Illustration 2. A weight, W, Fig. 64, rests upon the inclined 
plane AB. Find d so that P, the force which will just cause W 
to move up the plane, shall be a minimum. 

The normal pressure, N, between the weight and the plane AB 



168 CALCULUS I §91 

is equal to Tf cos (3 - P sin 6. Then F, the force of friction, is 
equal to 

H iW cos iS - P sin 6), 

where /t is the coefficient of friction. We have then 

F = m(^ cos |8 -Psin0). 

Since the force of friction must balance the components of P 
and W parallel to the plane AB, we have 

H{W cos i3 - P sin d) = (P cos e -W sin /3). 

Hence 

Win cos j8 + sin /3) 

— ^^- p ^-^ = cos + /i sm 6, 

or 

Tr(^cos^ + sin/?) ^^^^-^^^^^^_^^ (3^ 

where a is the acute angle whose tangent is n. Thus, the left- 
hand side of (3) is a maximum, and P a minimum, when d is acute 
and equal to tan~^ ju. 

„, . . . , f p • .K Tr(MCos^ + sini8) 
The mmimum value of P is then . 

Exercise 

In Fig. 64, find the minimum force, P, and the angle between its 
line of action and AB, which will just prevent the weight W from slid- 
ing down the plane. 

91. The Function mx + -y/a^ — x^. Frequently problems in 
maxima and minima lead to functions of the form mx ± ■s/a'^ — x^. 
The curve for 

y = mx ± y/a^ — x"^ (1) 

can be obtained by shearing the circle y = + \/a^ — x^ in the 
line y = mx. Every ordinate of the circle to the right of the F-axis 
is increased (or decreased if m is negative) by an amount propor- 
tional to the distance from the F-axis. To the left of the F-axis 
the ordinates are decreased if m is positive and increased if m is 
negative. 

The maximum value of y is easily found by placing 

X = a cos t. (2) 



§93] MAXIMA AND MINIMA 169 

Then from (1) 

y = a(m cos t + sin t) (3) 

= aVl + m^ cos (t — a), (4) 

where a = tan"^ — The maximum value of y occurs when 



X = a cos a = 



vTT 



92. Maxima and Minima by Limits of Curve. In case f(x, y) = 
is of the second degree in x and y, and in a few other cases, the 
maximum and minimum values of y can be found by determining 
when X changes from real to complex values. 

The method will be illustrated by an example. 
Let 

^ g^ + e 

^ ~ 2x + l" 
Then 

x = y ± V{y + S){y-2). (1) 

From equation (1) it is seen that for values of y greater than 2 
or less than — 3, x has two distinct real values. When y = 2 or 
— 3, X has two equal real values. When — 3 < ?/ < 2, a; is imagi- 
nary. This shows that the line y = c meets the curve in two 
distinct points if it is more than two units above or more than 
three units below the X-axis; that it is tangent to the curve 
when two units above and when three units below the X-axis, and 
that it does not cut the curve when it falls within the limits two 
units above and three units below the X-axis. Hence the func- 
tion has a minimum value 2 and a maximum value —3. 

Exercises 

1. Find the luaximum and minimum values of 

x" - 2x + 19 
2z +5 

2. Find the maximum rectangle which can be inscribed in a circle 
of radius 10. 

93. Maxima and Minima Determined by the Derivative. The 
first derivative has been used to determine the value of the argu- 



170 CALCULUS [§93 

ment corresponding to maxima and minima of functions. Im- 
mediately to the left of a maximum point the function is increas- 
ing with X and consequently the first derivative is positive. On 
the other hand, immediately to the right of such a point the func- 
tion is decreasing as x increases and the first derivative is negative. 
Similarly it follows that the first derivative is negative im- 
mediately to the left and positive immediately to the right of a 
minimum point. In both cases the first derivative changes sign 
as the independent variable passes through the value for which 
the function has a maximum or a minimum value. This change 
of sign may take place in a number of ways. 
Illustration 1. Thus, in the case of the function 

y = x^ -2x + 7, 
the derivative, 

I = 2. - 2 . 2(. - 1), 

is negative to the left and positive to the right of the line x = 1, 

dy 
When a; = 1, ^ = and the curve has a horizontal tangent. In 

the vicinity of this point the curve has the 
shape shown in Fig. 65. 

At first thought it might appear that if the 
first derivative is negative to the left and 
positive to the right of a certain point, it 
certainly must become zero at this point. 
This is, however, by no means the case, as 
the next illustration will show. 

Illustration 2. y = 4 + (x — 1)^. Although the minimum 
value of this function can be determined at once by noting that 
it represents a semi-cubical parabola with its vertex at (1, 4), the 
problem will be worked by the method of the calculus for illustra- 
tive purposes. 
The derivative, 

dy^ 2 ^ 

^^ 3(x-l)*' 

is negative when x < 1 and positive when x > 1. Hence the 
function is decreasing to the left and increasing to the right of 




§93] 



MAXIMA AND MINIMA 



171 



a; = 1. When x = 1, ?/ = 4. This value is a minimum value of 
the function. For z = I the derivative does not exist, as the 
denominator becomes zero. Let us see what really happens in 

the vicinity of x = 1. As x approaches 1 from the left, j- takes 

on larger and larger negative values. The form of the curve to 
the left and in the immediate vicinity of the point (1, 4) is some- 
thing like that shown in Fig. 66. The line x = 1 is tangent to 
the curve at this point. 

As X approaches 1 from the right, i.e., through decreasing values, 
the derivative becomes larger and larger. The form of the curve 
to the right of the line x = 1 is also shown in Fig. 66. The line 
X = 1 is also tangent to the portion of the curve obtained by allow- 
ing X to approach 1 from the right. 





Fig. C6. 



Fig. 67. 



It is now apparent that the first derivative may change sign 
without passing through zero. In the above illustration it changes 
sign by becoming infinite. 

The first derivative may change sign in still another way as 
illustrated by the curve of Fig. 67. Let us suppose that the de- 
rivative approaches — 1 as x approaches a from the left, and the 
value +1 as X approaches a from the right. The function has a 
minimum value at the point P, for the derivative changes from 
minus to plus as x increases through the value a and consequently 
the function is decreasing to the left and increasing to the right 
of X = a. 

The essential thing at a minimum point is that the derivative 
changes sign from minus to plus, and at a maximum point that it 
changes sign from plus to minus. 



172 CALCULUS [§95 

A derivative which is continuous at a maximum or a minimum 
point changes sign by passing through zero. But it may change 
sign by becoming infinite, as the second illustration shows, or by 
becoming otherwise discontinuous as explained above. This last 
type is of rare occurrence and will not be referred to again. 

Illiistration 3. y = x^ -^ 3. The derivative of this function, 

-p = 3x^, is positive for all values of x except x = 0, when it is 

zero. The function is increasing for all these values of x. At 
this point, (0, 3), there is a horizontal tangent but the function 
has neither a maximum nor a minimum at the point, for it in- 
creases up to the value 3 for x = and then continues to increase 
to the right. This illustration brings out clearly the fact that 
there is no reason for assuming that a function has a maximum or 
a minimum value at a point where the first derivative is zero. 
What kind of a point is the point (0, 3) ? 

94. Second-Derivative Test for Maxima and Minima. In the 
first of the three types of maximum or minimum points considered 
in §93, the first derivative changes continuously from positive to 
negative values or vice versa. For a maximum point of this type 
the curve is concave downward and the second derivative is 
negative at such a point. For a minimum point the curve is 
concave upward and the second derivative is positive. A con- 
venient test for the behavior of a function at a point where the 
first derivative is zero is then, to substitute the abscissa of this 
point in the expression for the second derivative. If the second 
derivative is positive the point is a minimum point; if negative, 
a maximum point. If the second derivative is zero, the test 
fails. This test also fails for maximum or minimum points where 
the first derivative is discontinuous. 

Examine the curves 

y = x», 

y = X*, 



95. Study of a Function by Means of its Derivatives. The 

following is a summary of the application of the first and second 
derivatives to tracing the curve representing a function: 



§95] 



MAXIMA AND MINIMA 



173 



1. The function is increasing if the first derivative is positive, 
and decreasing if it is negative. 

2. To find maximum and minimum points find the values 
of X for which the first derivative becomes zero or infinite. If 
the derivative changes sign at any of these points, the correspond- 
ing point is a maximum or minimum point according as the 
change is from plus to minus or vice versa. 

Points at which the first derivative is equal to zero can also bo 
tested by substituting the abscissa of the points in the second 
derivative. If the second derivative is positive, the point is a 
minimum point, if negative, a maximum point. 

3. Points of inflection are found by determining where the 
second derivative changes sign. As in the case of the first 
derivative, the change in sign can take place through zero or 
infinity. If the change is from positive to negative values the 
curve changes from being concave upward to being concave 
downward. 

The abscissas of the points at which the first and the second 
derivatives become zero or infinite we shall call the critical 
values. These values and these alone need be tested in study- 
ing the behavior of an ordinary curve. The investigation of a 
curve by means of its derivatives can be put in the tabulated 
form shown in the following illustrative examples: 



1. y 



(See Fig. 33.) 

dy 

dx 

d^ 

dx^ 



— X. 



dy 
dx 



= when x = 0. 



-1-^ = when x 



X 


dhj 
dx^ 


dy 
dx 


y 


X <0 
X > 


<0 
>0 


>0 

>o 


Concave downward, increasing 
Concave upward, increasing 



174 



CALCULUS 



^96 



Here (0, 0) is a point of inflection. There is neither a maxi- 
mum nor a minimum point. 
2. 1/ = ix3 - x^ + Ix + 2. (See Fig. 34.) 

g = ix^ - 2x + I 



= X 



i(:r - Dix - 3). 
2. 



dv 

-^ = when x 

ax 






= when x 



1,3. 
2. 



X 




dy 
dx 


y 


X < 2 
X > 2 


<o 

> 


Decreasing 
Increasing 


Concave downward 
Concave upward 


X < 1 

1 <x < 3 

X > 3 




> 
< 

> 


Increasing 
Decreasing 
Increasing 


1 
2 
3 







8 
3 

i 

2 
2 



(2, I) Point of inflection. 
(1, f) Maximum point. 
(3, 2) Minimum point. 

Apply second derivative test for x = 1 and x = 3. 

96. Applications of Maxima and Minima. In solving problems 
involving maxima and minima the first step is to set up from the 
conditions of the problem the function whose maximum or mini- 
mum value is sought. Frequently the function will be expressed, 
at first, in terms of two or more variables. Usually, however, 
there is a relation between these variables, and the function can be 



§96] 



MAXIMA AND MINIMA 



175 



expressed in terms of a single variable. After this has been done 
the maximum or minimum values can be found. 



Exercises 

1. Equal squares are cut from the comers of a rectangular piece of 
tin 30 by 20 inches. The rectangular projections are then turned up 
forming the sides of an open box. Find the size of the squares cut out 
if the volume of the box is a maximum. 

2. A man who is in a boat 3 miles from the nearest point, A, of a 
straight shore wishes to reach, in the shortest possible time, a point 
B on the shore which is 6 miles from A. Find the point of the shore 
toward which he should row, if he can row at the rate of 3 miles per 
hour and walk at the rate of 5 miles per hour. 

3. The horizontal component of the tension in the guy wire BC, 
Fig. 68, is to balance the horizontal pull P. If the strength of the wire 
varies as its cross section, and if its cost varies as its weight, find the 
angle such that the cost of the guy wire shall be a minimum. 

B 




Fig. 68. 



Fig. 69. 



4. Find the length of the shortest beam that can be used to brace 
a wall if the beam passes over a second wall 6 feet high and 8 feet from 
the first. 

5. A steel girder 30 feet long is moved on rollers along a passageway 
10 feet wide, and through the door AB, Fig. 69, at the end of the pas- 
sageway. Neglecting the width of the girder, how wide must the 
door be in order to allow the girder to pass through? 

6. A sign 10 feet high is fastened to the side of a building so that the 
lower edge is 25 feet from the ground. How far from the building 
should an observer on the ground stand in order that he may see the 
sign to the best advantage, i.e., in order that the angle at his eye sub- 
tended by the sign may be the greatest possible? The observer's eye 
is 5 1 feet from the ground. 

7. A man in a launch is m miles from the nearest point A of a 
straight shore. Toward what point on the shore should he head his 
boat in order to reach, in the shortest possible time, an inland point 



176 



CALCULUS 



[§96 



whose distance from the nearest point B of the shore is n miles? The 
man can run the boat Vi miles per hour and can walk Vj miles per hour. 
The distance AB is p miles. 
Ans. Toward a point such that 

sin 61 sin dt 



Vl 



V2 



where 0i and 62 are the angles made by the paths of the man with the 
normal to the beach. It will be noticed that the path taken by the 
man is similar to that followed by a ray of light in passing from one 
medium to another with a different index of refraction. 




-.25 



.5 1 1.5 

Fig. 70. 

8. A man in a launch is m miles from the nearest point A of a 
straight shore. He wishes to touch shore and reach, in the shortest 
possible time, a second point on the lake whose distance from the 
nearest point B on shore is n miles. In what direction must he head 
his boat if the distance AB is p miles? 

The path taken by the man is similar to the path of a ray of light 
reflected by a plane surface. 

9. It is desired to make a gutter, whose cross section shall be a 



MAXIMA AND MINIMA 



177 



segment of a circle, by bending a strip of tin of width a. Find the 
radius of the cross section of maximum carrying capacity. 

10. A sector is cut from a circular piece of tin. The cut edges of 
the remaining portion of the sheet are then brought together to form a 
cone. Find the angle of the sector to be cut out in order that the 
volume of the cone shall be a maximum. 

11. The stiffness of a rectangular beam varies as its breadth and as 
the cube of its depth. Find the dimensions of the stiffest beam which 
can be cut from a circular log 12 inches in diameter. 

12. The strength of a rectangular beam varies as its breadth and as 
the square of its depth. Find the dimensions of the strongest beam 
which can be cut from a circular log 12 inches in diameter. 

1.5 



L26 



.75 
(4) 



.25 



k^^ 










3^(2) 




(1) 2/ = e"* +0.56-^"* 

(2) y=e- 

(3) 2/ = e ^-o.ie 

(6) 2/ = e-* -1.56 ^ 


>A(3) 




p^ 


\, 








|(6) 


\ 


s^ 














— - 



1 1.5 

Fig. 71. 



2.6 



13. Consider the sum 

y = ox" + bx*" 
for positive values of x only. First, if n and r are of like sign, show 
that: (1) a maximum or a minimum value exists if a and b are of 
unlike sign; (2) neither a maximum nor a minimum value exists if a 
and b are of like sign. Second, discuss the same cases if n and r have 
opposite signs. 

14. Determine the exact values of the maxima shown in Figs. 70 
and 71. 

Hint. Consider first the general case 



12 



y 



e~' — ae 



CHAPTER XI 

POLAR COORDINATES 

97. Direction of Curve in Polar Coordinates. Let BPQ, 
Fig. 72, be a curve referred to as pole and OA as polar axis. 
Let P be any point of the curve and let PT be a tangent to the 
curve at this point. Let PS be the radius vector of the point P, 
produced. 
A point describing the curve, when at-P, moves in the direction 
PT. This direction is given by the 
angle ^ through which the radius 
vector produced must rotate in a posi- 
tive direction about P, in order to be- 
come coincident with the tangent line. 
An expression for tan^ will now be 
found. Let Q, Fig. 72, be a second 
point of the curve. PR is perpendicu- 
lar to OQ, and PM is a circular arc with as center and radius 
OP = p. 

tan^= iJ^otani2QP= iJS'o ^. (D 

The infinitesimals PR and RQ can be replaced by PM and MQ, 
respectively, if (see §60) 

lim PR 

A0=O PM 




Fig. 72. 



1 



(2) 



and 



= 1. 



lim RQ 

Equation (2) is true by equation (3), §56 
(3) follows: 

lim RQ^ 

Afl-.0 MQ 



(3) 
The proof of equation 

lim RM + MQ 



Ae=o MQ 

Urn p(l -cos Ag) + A p 

A9=0 ^p 

, , lim p(l-cosA0) Ae 
178 



Ad 



Ap 



§97] 
Hence 

since 



POLAR COORDINATES 



179 



lim RQ 



= 1. 



Ae=o MQ 
lim 1 — cos Ad 

A9=0 ^0 

From (1), (2), and (3) it follows that 



= 0. 



, , lim PM 



Hence 



lim pAg 



tan^ 



dp 



tan^ = ^- 
dd 



(4) 



This formula* can be easily remembered if the sides of the tri- 
angular figure MQP, Fig. 72, are thought of as straight lines, and 




d sin 
' This formula enables us to give another proof for — -yz — . In polar codrdinates 

p = sin 8 represents a circle, Fig. 73. By geometry, x// = e. Then 

p sin 

tan^ = tanfl=^ - ^. 

de 



dp 

de "■ 

d sin 

de 



de 



cos 8, 



180 CALCULUS [§97 

the angle MQP as equal to \p. Then the tangent of rp would be 



de p_ 
dp 
dd 



dp dp 



dv 
Formula (4) corresponds io -r = tan t in rectangular coordi- 
nates. 

Illustration L If p = e"*, 

dd = «^ 

and tan \I/ = -, a, constant. 

^ a' 

Illustration 2. Find the equation of the family of curves for 
which the angle between the radius vector produced and the 
tangent line is a constant. 

tan yp = k. 



or 



Integrating 



or 



dp 
dd 


k, 


dp 


; 


dd _ 
P 


1 

k 


dp 
P 


>• 


IP = 


1-- 


P = 


«:— 


= 


e e*, 


P = 


Kek, 



where K is an arbitrary constant. 



^98] POLAR COORDINATES 181 





Exercises 


Find tan \p for each of the following curves : 


1 '^. 


4. p = a(l — cos e). 
5 o 


2. p = ad. 


'' 1 - cos » 


3. p = e"'. 


6. p = o cos (0 — a). 



98. Differential of Arc: Polar Coordinates. We. shall now 

ds 
find an expression for -v^ in polar coordinates. From Fig. 72, 



dd 
(chord PQ) 2 = (PRy + (RQ^. 



From which 



lira /chord PQV lira /PRy , lira /RQV 

Replacing chord PQ by arc PQ = As, PR by PM = pA0, and 
RQ by MQ = Ap, 



(1) 
(2) 

(3) 

It corresponds to (ds)^ = (da-)^ + (%)'' in rectangular coordinates. 
It can be remembered easily by the help of the triangle MQP, 
Fig. 72. 

The length of the curve can be expressed as a definite integral. 
Thus: (See Fig. 74) 

lim -^ _,^ 

s = A« = 2Lf P^ 





lira / 

A9 = 1 




Y lim fpAey , 1 


Therefore 






[de) = ^ + [de) 


and 










dS = Jp2+(^P)W 


This formula can 


be written 








(ds)2 = p2(d0)2 + (dp)^ 



182 CALCULUS §98] 

= a/™o % ViPRV + {RQy 



B 
= /3 



-J'i-0 D'VQV (ff)'- 



= a 

= fi 






Illustration. Find the entire length of the curve p = a(l — cos 0). 
This curve is symmetrical with respect to the polar axis. The 
length of the upper half will be found and multiplied by 2. 

dp 



= j Va\i - cos ey + a2 sin2 d dd 



=^^de 



a 

sin -H" rf0 



= — 4a cos — = 4a. 
^ lo 
s = 8a. 

Exercises 

1. Find the entire length of the curve p = 2a sin 9. 

2. Find the entire length of the curve p = a(l — sin 6). 

a 

3. Find the entire length of the curve p = a sin' -^• 

o 

4. Find the length of p = e°* between the points corresponding to 
6=0 and = v. Also between the points corresponding to = 

and e =Z- 



§99] 



POLAR COORDINATES 



183 



5. Prove formula (3) directly from 



and 



X = p cos d, 
y = p sin e, 

ds = Vidxy + (dj/)2. 



99. Area : Polar Coordinates. Find the area bounded by the 
curve p = fid) and the radii vectores 6 
= a and 6 = ^. We seek the area 
BOC, Fig. 74. Draw radii vectores 
dividing the angle BOC into n equal 
parts Ad. Let POQ be a tj'pical one 
of the n portions into which the area 
is divided by these radii. The angle 
POQ is Ad. The line OP makes an 
angle 6 with the initial line OA, and ^^°- '^' 

its length is p = f(d). Denote the area of BOC by A. 

lim 




o7 



^ =nT^XPOQ^ 



(1) 



Replace^ POQ by the circular sector POR whose area is ^p^'A^. 

Then 

^ s 



or 



'I 



P^dd, 



A = h\ [mv dd. 



Exercises 



(3) 



1. Find the area bounded by the curve p = 2o sin e. 

2. Find the area bounded by the cardioid p = 2a(l — cos 6). 

, 1 + cos 29 
Hint. cos*0 = ^ 



iLetA^ = OPQ (Fig. 74). 



PR and QS are arcs of circles. Then 
OPR < A^ < OSQ, 



Jp«Ae <AA< i(p + Ap)'AJ. 



184 CALCULUS [§99 

3. Find the area bounded by p = 2a(l + sin e). 

4. Find the area bounded by one loop of p = 10 cos 2d. 

5. Find the area bounded by one loop of p = 10 sin 26, 

6. Find the area bounded by one loop of p = a cos Sd. 

7. Find the area bounded by one loop of p^ = 10 cos 29. 

8. Find the area swept out by the radius vector of the curve 
p = ad, as varies from to 2ir. 

9. Find the area bounded by the radii vectores = 2, ^ = ir, and 

the curve p = -r^- 

10. Find the area bounded by the radii vectores = 0, 6 = ^ and 
the curve p = 5^^. 



CHAPTER XII 

INTEGRATION 

100. Formulas. In Chapters III, VI and VII the following 
formulas of integration, with the exception of (19), have been used. 
They are collected here foj reference, and should be memorized by 
the student. 

1. fw du = — 7-r w»+i + C, if w 7^ - 1. 
J n + 1 



3. J e» rfw = e» + C. 

4. I a" du = , a» + C. 

J log, a 

5. j sin udu = — cos u -\- C. 

6. J cos w dw = sin w + C. 

7. I sec'^ udu = tan u + C. 

8. I csc'^ udu = — cot w + C. 

9. I sec u tan udu = sec m + C 

10. I esc M cot udu = — esc w + C 

11. J tan udu = log sec w + C. 

12. J cot udu = log sin w + C 

13. J sec udu = log (sec w + tan u) + C 

14. { CSC udu = — log (esc w + cot u) + C. 

15. I —r = sm-i - + C. 



185 



186 




CALCULUS 


16. 


du 
a^ + u^ 


= - tan-i - + C. 
a a 



[§100 



17. I .- = - sec-i - + C. 

J u\/u'^ — a2 « « 

18. I ,J^ = log (m + Vu^ ± a^) + C. 



r du ^ 1^ 

J u^-a^~ 2a 



19. Iri^.-^loK^ + C.ifX- 

= K- log — j \- C,itu < a. 

2a ^ a + u 

Formula (19) is proved as follows: 



.^^lU Ll 

u2 — qZ 2a \u — a u-\- aj 

f du ^ i_ rr_i 1 1 

J u^ - a^ 2aJ [u-a w + a J " 

_ 1 r du 1 r du 

~ 2a J u — a 2a J u-\- a 

= 2" log {u - a) - 2^ log (w + a) -f C. 

1 , u-a ^ 

= ^ log — , h C. 

2a ^ u + a 

This formula leads to the logarithm of a negative number if m < a. 
To obtain a formula for this case write 

^_ = ir__i Li. 

u^ — a^ 2a L a — u a-\-uJ 
Then 



/ 



du 1 , a — M . _ 

= o:; log :,-n-7: + C- 



■u^ — a^ 2a ^ a + w 
Exercises 



r xdx r 

* J Vie - z2* ^* J V^ 

^* J Vie - X*" *' J ^' + 16 



2-16 

dx 



§100] 



INTEGRATION 



187 



r dx 

• J16-X 

I 



16 



xdx 
x^ - 16* 

dx 



xy/x^ - 16 

9. fcot 1i dt. 

r {x+ a)dx ^ 
J ^' + 2ax 

11. fix^ - 16)^ X dx. 

12. fsin (2x - 3) dx. 

13. fsec^ (5a + 2) da. 

25. I e2 +e~2 dx. 

26. la^'' dx. 

27. r(2x + 4)''-" dx 

^^' J 7x* + ll 
30. I 



4. Jsec (2(? + 4) tan (20 + 4) de. 

5. fcsc^ (3 - 2</.) d<^. 

7. fe^^'^^ainede. 

C 31^ dt 
J 30<'+ 13' 

9. JiVc - Vxy dx. 

20. J \/3 + 4x dx. 

/I 
e^dy. 

23. fe*'*" (2^ +2) sec2(2x + 3) dx. 



dx. Divide numerator by denominator. 
31. le^'dx. 



dx 



/ dx 
4x2 _ 

/ 

J Vie - 9x2" 

35. ftan (3a + 4) da. 



33 



34 



9 

cos X dx 
4 + 3 sin X 

dx 



cos" (3x - 2) 
36. Htan e + cot ey dd = tan e - cotB + C. 

39. Jcos (3< - 4) dt. 

„ 40 r ^*- 

38. J cot (5< - 8) dt. J ^ - *2/' 



/■ 



37. I (sin -z — cos 50) dd. 



188 CALCULUS [§100 



dz 61. fcos^ (3x - 2) sin (3x - 2) dx. 

62. I sec* (9 - 7x) dx. 



63. I e 3 sec^^dx. 



•{. 63. 

43. J tan (2x - 5) dx. J 

44. Jsec (2y + 4) dt/. 64. J tan* x sec* a: dx. 
46. fcsc (2t/ — 7) dy. 65. J cos* 3x sin 3a; dz. 

46. J cot (3< + 11) d/. 66. ftan^ 5x sec* 5xdx. 

47. I sec* (^ — 5]dx. 67. I sec* x tan xdx. 

48. jTcos (3 - 2x) dx. «8. J csc^ x cot x dx. 

Aa r 2x + 5 ^ 69. fx tan (2x* - 5) dx. 
**• I x* + 5x + 4r^- % 

/ _ I sin X cos X dx 

62 



/ <^^ 72. (e* —^dx. 

V9<* - 4 J * 

/ <^^ 73. fiJ - x»)» dx. 
V4-9/*- 

J <\/9<* - 4 "^ 

/ tdt 75. ft^Qt^ - 17 dt. 

V9i^^4: r ^dx 

r tdt ^^J 

j9t*-4 n 

. I sec 5x dx. ' j i 



V9 -X* 
sin 5x dx 



57. I sec 5x dx. ' / ^ ^*^^ ^^ "I" ^^ 

68. fsin (cot + a) dt. 78. fe^''' ^^ cos 3x dx. 

69. Jcos* 4x sin 4x dx. 79. fe^' + <^ + '^{x + 3) dx. 
60. j sin* (x +3) cos (x + 3) dx. 80. | cos 5x sin 3x dx. 



INTEGRATION 

83. Jsin 3t cos 4t dt. 



84. I sin«5«d<. 



§100] INTEGRATION 189 

81. I cos 3x cos 5x dx. 

82. I sin 7x sin 4x dx. 

r 

85. I sin mt cos n< di, where m and n are integers. What is the 
value if m = n? 

86. I cos (3w< + a) sin (3w< + a)dt. 



'I? 



88 



89 



90 



2x + 3 , 

x^ + 3x^+7 
x* + 9 



•J 3^^^^- ■ 

/3x-f-2 , 

91. fsin* 5fl cos 56 dO. 

92. I sin^ X dx. 

93. J xVie - x2 dx. 

94. I sec2 (l + 2J dx. 

95.jy^^dt. 

96. I sec 2 4x tan 4x dx. 

97. I \ dx. 

109. fsec {3<l> - 2) tan (30 - 2) d,/.. 

110. rtanH2x - 1) sec* C2x - 1) dx. 






dx 



V5 - 7x2 
dx 



VSx^ - 5 

00. I sin 4x cos 6x dx. 

01. f(V^-V^)'dx. 
'2x + 3 






2x + 7 
dx 



dx. 



V3x +2 

04. I sin* 2x cos 2x dx. 

05. I Vsinx cos x dx. 

06. fe-^dt. 

07. r(2x - 5)^dx. 



08. 



/ 



sec odx. 



190 CALCULUS [§101 

"^- j l + tan3x ^^- "8-j3^M^4 

112. fv2^^3idx. 119. [ —y^ 

•^ J yVsy 



dy 



113. ftan (5 - 2x) dx. f* vdv 

J 120 I —^=^ 



114 
116 



f .i. J Vs.' -7 

J 5 - 3x2 ]L21. I sec^ tan e dd. 

• J ¥~^7 "2- J VlP^' 

116. I^^dx. 123. |sm'|cos|dx. 

/x + 4 C X 

— ; ?; <ix. 124. I cos 2xsin o<ia;. 

x^ - 9 J 2 

101. Integration of Expressions Containing ax^ + bx + c, by 
Completing the Square. 
Illustration 1. 

/ dx ^ r dx 1_ J X + 2 

X2 + 4X + 9 J(x + 2)2+5 VS VS 

Illustration 2. 

/rfx r dx C dx 

V3 + 4x -"x^ ~ J VS - (x2 - 4x) ~ J \/3 + 4-(x-2)2 
dx . X — 2 



■J: 



\/7 - (x - 2)2 \/7 

Illustration 3. 



+ C. 



r dx _ 1 r 

J7x2 + 3x + ll IJ 



x^ + ^x + tEit + V - ~lU 
dx 



1 14 ^ _, x + -A- , ^ 

s — ;== tan * .> . + C 

' V^99 V^99^ 

14 



2 



tan 



-. lii+3 + c. 



299 \/299 



§101] INTEGRATION 191 

Illustration 4. 

dx 



r dx i_ r dx 1^ r 

J VQ+2x-3x^~ V3J V2+lx-x^~VSJ V2-{x^-l^ 

1 C dx 1 . _. x-l 

= — 7= I — ■— = — -7= sin ■* — 7= + C 

\/3j VY -ix-\r \/3 VY 



— ;- Sin^ ;p:r— + C. 

V3 \/l9 



Exercises 



J x^ + 6x +25 J 



^- J x2 - 6x + 5 J 5x^ - 

4. f-^^ 9. f 

J A/2a;2 + 2x - 3 J 



V2 + 3t - 2t2 
dx 



8x + l 
dt 



Vl +2t + 2t2 



2x2 _|. 5x - 3 



,„ r 2x-5 ^ f 2x+6-ll _, 

1°- Jx^ + 6x+25^^ =J^2+6x+25'^^ 

/ (2x + 6)dx _ r___^5___ 
x2 + 6x + 25 J x2 + 6x + 25* 

J V2x2^+ 2x - 3 J VS - 4x - x2 

/4x + 1 1 C 

^^-p2^-dx. 14. J- 

16 r ^^ 18. f ^^ 

' J xV3l^ 6x + 5x2 J V4J/2 + 122/ - 7 

17. f-^-=' 19. fg^- 



dx 1 

15. I ^ Substitute x = -• 

xv/2x2 + 3x - 2 ^ 



(Zx 

24x -9 



192 CALCULUS [§102 

20. I 23. I . ^ =' 

J V8 + 12x - 4x» J x\/^+12x - 7x2 

J \/8 + 12x - 4x2 J a 



X VSx^ + 12x - 4 



/ 



22. 



16x2 - 24x + 24 



102. Integrals Containing Fractional Powers of x or of a + bx. 
Illustration 1. 



/: 



x^ — x^ 
— ax. 



3 +4 
Let X = 2*. Then dx = 62^ dz, and 

J ^i + 4 ^ j0^ + 4' ^" J^^ + 4 '''• 

The integration can be performed after dividing the numerator 
by the denominator until the degree of the remainder is less 
than 2. After integration replace z by x^. 
Illustration 2. 

\x + 2)^ + 4 



/! 



dx. 



(x + 2)^-3 
Let X + 2 = z<. Then 

r(x + 2)U4 f(?!+-41^'rf. 

Divide the numerator by the denominator. The integration can 
readily be performed. After integration replace 2 by (x + 2)*. 

In general if fractional powers of a single linear expressior, 
a + bx, occur under the integral sign, let a -{- bx = z", where n is 
the least common denominator of the exponents of a + bx. The 
linear expression a + bx reduces to x when a = and 6=1. 
See, for example, Illustration 1. 

Exercises 



§103] INTEGRATION 193 




f. 
■■/' 



^„ , 3x- 2 , 

10. I . dx. 

x\/2x + 3 



(a;i + a;3) dx. 



12. I . dx. 

Sx\/2x - 3 



14. 



J 1 + (X + 2)^ 



13. I v-^-T--^- -rx ^^^ 
1 + (X + 2) 
Vx — 3 dx 



P 



x + 4 

x(l + x^) 



15. I ~" ' T dx. 

jg rV2x + 3 dx. 



3x - 2 

103. Integrals of Powers of Trigonometric Functions. 

(a) I sin" X cos" x dx where at least one of the exponents is an 

odd positive integer. This includes I sin"» x dx and J cos" x dx 
where the exponents are odd. 
Illustration 1. 

J sin^ X cos^ X dx = J (1 — cos^ x) cos* x sin x dx 

= J cos* xsinx dx — J cos* xsinxdx 
_ cos' a; cos^x _, 
- 3- + -5— + ^- 

Illustration 2. 

I cos' X dx = 1(1— sin^ x) cos x dx 

= I cos X dx — j sin* x cos x dx 

sin'x , _ 
= sin X — — o r C. 

It is seen that the process consists in combining one of the func- 
tions sin X or cos x with dx to form the differential of — cos x or 
of sin X, respectively, and of expressing the remaining factors 
of the function to be integrated in terms of cos x or sin z, 
respectively. 

13 



4. I sin* X dx 

6. I \/sin X cos' x dx. 



194 CALCULUS [§103 

Exercises 

1. fsin'zdx. f cos»a; 

•7. 7. -y== dx. 

2. j sin" X COS* z dr. J ^^^'^ ^ 

3. I cos* X sin' x dx. g I ^'"' ^ ^^^ 
r . . . t/ (cos^)' 

9. I sin* a cos' a da. 
6. fcos* X sin' x dx. 10- J cos^ (2x+3) sin' (2x+3) dx. 

(b) I sin" z COS" x dx when w and n are both even positive 

integers- In this case make use of the relations: 
sin'^ X = 5(1 — cos 2.t). 
cos^ X = 5(1 + cos 2x). 
sin X cos X = ^ sin 2x. 
Illustration 1. 

I sin' X dx = H(l ~ cos 2x) dx = \\dx — \\ cos 2x dx. 
_ X sin 2x 

~2 r^ + ^- 

Illustration 2. 

I sin' X cos' X dx = 5 j sin' 2x dx = | 1 (1 — cos 4x) dx 

_ X sin 4x 
~ 8 32~ "^ 

Illustration 3. 

J cos* X dx = J I (1 + cos 2x)' dx 

= ij(l 4- 2 cos 2x + cos' 2x) dx 

= ix + 4 sin 2x + I I (1 + cos 4x) dx 
= f ^ + 4 sin 2x + -^2 sin 4x + C. 
Illustration 4. 

J sin' X cos* X dx = J (sin x cos x)* cos* x dx 

= I J sin' 2x (1 + cos 2x) dx 

= iV J (1 — COS 4x) dx + I I sin' 2x cos 2x dx 
= iVx — Vv sin 4x + Vif sin' 2x + C. 



§103] INTEGRATION 195 

Exercises 

1. js'm'^xdx. 4. jsm*xdx. 

2. J cos* 2x dx. 5. fsin* 3x dx. 

3. / sin* X cos'' x dx. 6. j cos* 5a; dx. 

(c) I tan" a; rfx and I cot" x da;. 
Illustration 1. 

J tan^x dx = J tan^x (sec^x — 1) dx = „ — ftan^ x dx 

tan' X f/ « . V , 

= — 5 I (sec'^x — 1) dx 

tan^x , , /^ 

= — o tan X -r X + C. 

Illustration 2. 
J cot' X dx = J (csc^ X — ly cot X dx 

= I esc* X cot X dx — 21 csc^ x cot x dx + I cot x dx 
= — J CSC* X + csc^ X + log sin x -\- C. 

(d) I sec X dx and J csc" x dx, n an even integer. 
Illustration 1. 

J sec* X dx = J (1 + tan^ x) sec'^ x dx = tan x + | tan' x + C. 
When n is odd this method fails. (See §106.) 

(c) / tan" z sec" x dx and J cot*" x csc" x dx when n is a positive 
even integer, or when m and n are both odd. 
Illustration 1. 

J tan* X sec* x dx = J tan* x(l + tan'^ x) sec^ x dx 
= I tan^x + 7 tan^x + C. 
Illustration 2. 

J tan' X sec' x dx = j tan'^ x sec^ x sec x tan x dx 

= I (sec'^ X — 1) sec^ x sec x tan x dx 
= I (sec* X — sec' x) sec x tan x dx 
= i sec' X — 5 sec' x + C. 



196 



CALCULUS 



I §104 



If m is even and n is odd the methods of §106 must be used, for 
the integral reduces in this case to the integral of odd powers 
of the secant. 

Exercises 

9. I tan^ X sec' x dx. 
LO. I tan' X sec* x dx. 



1. I tan* X dx. 

2. 1 CSC* X dx. 

3. 1 tan^ X sec' x dx. 

4. I tan* X sec' x dx. 
6. I cot* X dx. 

6. I csc« X dx. 

7. I tan* X sec'* x dx. 

8. I sec* X dx. 



I tan 5 z sec* a; dz. 
2. I (tan 2 x + tan* x) dx. 
1. I sec' X tan^ x dz. 

4. fctan e -f cot OY dd. 

5. jtan^ede. 
. I sec' e tan~* 6 d9. 

104. Integration of Expressions Containing s/a.^ — x% Va* + x*> 
\/x2 — a^ by Trigonometric Substitution. The methods of §103 
find frequent application in the integration of expressions which 
result from the substitution of a trigonometric function for x 
in integrals containing radicals reducible to one of the forms 
Va* + x\ Va* - x\ or Vx* - a^. 

Illustration 1. I \/a^ — x^ dx. Let x = a sin 6. Then 
dx = a cos d0, and 

JVa^ - a;2 dx = Ja^ cos^ dO = la^{e + | sin 2d) + C 
= ^a2(0 + sip cog ^) + c 

= la' [sin-i ^ + ^, aA^":^] + C 

= W sin-' ^ + JxVa* - x^ + C. 
Illustration 2. j -\/a2 -|- a;2 ^3 dx. Let x = a tan 0. Then 
J Va» + x^x'dx =a5 J tan^ sec^ d9 

= a'^Jtan^ 6 sec^ tan sec d^ 
= a»J(sec* d - sec* 0) tan sec & dd 



§104] 



INTEGRATION 



197 



= a6(i sec* d - I sec3 6) + C 



a' 



1 + 



1 + 






Illustration 3. 



/ 



5 3 
5 3 



+ (7. 



dx. Let X = a sec 6. Then 



dx = a sec d tan d0, and 






sec5tan2 0d0 



sec 

= a Jtan2 6 dd 

= a (tan 9 - 0) + C 

= a-* /-; — 1 — a sec"^ - + (7 

= Vx^ — a2 — a sec-i - + C. 
o 

The integration can also be performed directly if the numerator 
is rationalized. Thus, 

rVx2 - a^ , 



(x2 — a^)dx 



■ r-7^£= - a^ f 



dx 



= \/x2 — a2 — a sec-i - + C. 
a 

The substitutions used in these illustrations are summarized in 
the following table: 



Radical 


Substitution 


Radical becomes 




X = a sin ^ 
X = a tan 9 
X = a sec d 


a cos 9 
a sec 9 
a tan 9 


Va' +x^ 
Vx^ - a^ 



198 CALCULUS [§104 



Expressions involving s/ax^ + 6x + c can frequently be inte- 
grated by completing the square under the radical sign and making 
a trigonometric substitution. 

Illustration 1. 



r xdx r xdx 

J V3 + 2x- x2 J V'4 - (x - 1 



Let X - 1 = 2 sin 9. Then a; = 1 + 2 sin ^ and dx = 2 cos 6 dd. 
Hence 



/ x dx r 

VS + 2x - a;2 "" ^ J 



(1 + 2 sin 6) cos Ode 



2 cos 
= J(l + 2 sin 0)d0 

= - 2 cos ^ + C 

X — 1 
= sin-» —2 VS + 2x - x^ + C. 

Illustration 2. 



0^]' 



/ rfx r rfx__ 
\/(2aa; - x^y ~ J [a^ - (x - a) 

Let X — a = a sin ^. Then x = a(l + sin d) and dx = a cos rf5. 

dx r a cos 5 

•\/"(2ax - x^y ~~ J a'coss 

= ^Jsec^ede 



= -„ tan + C 

^ 1 sin g 
a2 cos "^ ^ 
X — a 
1 ~^ 



a2 1 , _ 

- V2aa; - x^ 



+ C 



o'' \/2ox - x2 



§105] INTEGRATION 199 

Exercises 

dx 



dx 



1 r dx ^ r dx r 

2./>^'... »./;vfc, »-/a-.,vr^ 

3./xvrTT..x. 6.J^^7^,- »./^^;r~r» 

10. J (a^ - x^y-dx = 3a^f cos* sm^O de. 



Hint. Let 



S 3 . 

xs — a^ sin'' e, 



X = a sin^ 0. 



4)5 
dx 



(16 - x^)* 
dx 



r x^dx r dx r dx 

' J Va^~^^' ' J xWx^^^' ^- J a;(x» - 

12. fV9 - 5x2 dx. 14. f{9 ~x^)^dx. 16. 1 

J (x2 + 6x + 25) 2 J (x2 + 4x - 5)^ 

Vl^^^" 20. Jv'2 + 6x-x^cix. 

105. Change of Limits of Integration. In working the pre- 
ceding exercises by substitution it was necessary to express the 
result of integration in terms of the original variable. In the 
case of definite integrals this last transformation can be avoided 
by changing the limits of integration. 



L. I x^Vo^ 



Illustration 1. | x^ Vo^ — x^dx. Let x = a sin 6. Then 
dx= a cos d dd. 



When X = 0, sin = and ^ = 0. 

TT 

When X = a, sin ^ = 1 and ^ = o* 



200 CALCULUS [§105 

As X varies continuously from to o, varies continuously from 

TT 

to ^ • Hence we have 



I x^y/a^ - x^dz = a^\ "sin^ Q cos^ Q 
Jo Jo 



dd 



= a*{ld - A- sin Ad) 


2 

n 


wa* 
~ 16 


r** x' dx 
Illustration 2. 1 — -. Let x = a tan 6. Then 

Jo Vfl^ + x^ 


dx = a sec^ 6 dd. 


When X = 0, tan = and 6 = 0. 


■r,^, . /> - 1 /. 'T 





When X = a, tan = 1 and ^ = 7* 
As X varies continuously from to a, varies continuously from 

IT 

to T- Hence we have 
4 

J^* x'dx C* \~* 

-^=== = a' 1 tan» d sec d dd = a^i sec^ d - sec d)\ 
VS^T^^ Jo lo 

= ia\2 - V2). 

\/a2 — x^ dx. By using the substitution 
X = a sin we obtain 

/*2 

d5 



JVa^ - x2 dx = a^ j cos^ 
«/0 



= iaM^ + ^ sin 20)r 



Tra'' 



l2 

~ 4 * 

The above integral is of frequent occurrence in the application 
of the calculus. The integrand, \/a^ — x^, is represented graphic- 



§106] INTEGRATION 201 

ally by the ordinates of a circle of radius a, center at the origin. 
The integral then represents the area of one-quarter of this circle. 
(See §§64 and 65.) The value of any integral of this form may 
be written down at once. Thus, 



I V4 -{x- 5)2 dx = I \/4 - M^ rfw = 
Jo Jo 



7r22 



s: 



^^TT^ _ _ 7r(3-hz^) 

\/3 + z^ - a;2 dx = -. 



Exercises 



1. I (9 - a;*)* dx. 7. I 

Jo Jo (o^ + X 

,. f— i^. s. r 

Jo V2ax-x' J^A 

r x^dx g r_ 

• X V9^r^ ' Jo (^ 

,. r dx . ^, r 



Vi" X (x^ - 4)' 
dx 



+ 16)2 
dx 



x'^ix^ - 9)^ 



6 



I ^25 - x^dx. 11. I \/9 - (x - 4)2 dx. 

V^9 _ a;2 dx. 12. I Vb'' - 2/2 _ x2 dx. 

Jo 

106. Integration by Parts. The differential of the product of 
two functions u and v is 

d{uv) = udv ■}- V du. (1) 

Integrating we obtain 

uv = \udv -\- \vdu 
From which 

r u dv = uv — J V du. (2) 

This equation is known as the formula for integration by parts. 
It makes the integration of u dv depend upon the integration of 
d» and of v du. 



202 CALCULUS [§107 

Illustration 1. j x log x dx. Let log x = u and x dx = dv. 
The application of (2) gives 

I xlogx dx = Ix^ log X — I I x^- dx 

= |x2 log X - lx^ + C. 

Illustration 2. I xe'*da;. Let e^'dx = dv and x = u. The 
application of (2) gives 

I xe^'dx = \xe^' — \ \ e^' dx 
= |xe3' - W + C 
= ie3'(3x - 1) + C. 

If we had let xdx = dv and e'* = w we should have obtained a 
more complicated expression to integrate than that with which 
we started. 

Exercises 

1. I x* log X dx. 2. j X cos x dx. 3. I sin~i x dx 

4. I x'' e*' dx. (Apply formula (2) twice in succession.) 

6. I tan~i X dx. 9. I x sin' x dx . 

6. I X sin X dx. 10. I log x dx. 

7. I x" log X dx. 11. I x^ sin 2x dx. 

8. I x'^ tan~^ 2x dx. 12. I sin x log cos x dx. 

107. The Integrals J e" sin nx dx, J e** cos nx dx. Let 

M = sin nx and dv = €"dx. Then 

Jc* sin nx dx = -e»* sin nx ( e"' cos nx dx. 
a a J 

A second integration by parts with u = cos nx and dy = e"dx 
gives 

f . , 1 . n w* f . . 

I e"sm nxdx = -e<»*sin nx ;e"cos nx x I e" sm nx dx. 

J a a^ a^ J 

The last term is equal to the integral in the first member multiplied 



§107] INTEGRATION 203 



by -^- On transposing this term to the first member we obtain 

^ — I e"^ sm nx ax = ^ (a sm nx — n cos nx) + C. 

Then 

/©ax 
e sin nx dx = —^-r — i (a sin nx — n cos nx) + C 

eax 

sin (nx - a) + C, (1) 



where 



and 



cos a = 



sm a 



Va^ + n2 
q 



The student will show in a similar way that 

/gax 
e"cos nx dx = ~ir~, — o (n sin nx + a cos nx) + C 
a2 _i- n2 V 



Va^ + n2 
where 



a^ + n^ 
eaz 

cos (nx - a) + C, (2) 



cos a = 



and 

n 



sm a = 



Va^ 



Exercises 

The student will work exercises 1-5 by the method used in obtaining 
(1) and (2) above. In the remaining exercises he may obtain the 
results by substituting in (1) and (2) as formulas. 

1. fe-" sin 7t dt. 6. \e-'^ cos bt dt. 

2. fe-^ cos St dt. 7. je-"-" sin ut dt. 

3. fe-o" sin St dt. 8. fc-o-^ cos co< dt. 

4. fe-o-^' cos 4< d<. 9. fe-"'" cos 5t dt. 

5. je~''smxdx. 10. j e'"-' s'm 4:t dt. 
11. Find a in exorcises 1-10. 



204 CALCULUS [§109 

108. J sec^ X dx. This integral can be evaluated by a method 
similar to that used in the last article. 

/sec...x=/sec.sec..<*. 

= sec X tan x — \ sec x tan^ x dx. 

Since tan^ x = sec'^ a; — 1, 

I sec' xdx = sec x tan ^ — j sec' x dx + j sec x dx. 

Transposing the next to the last term to the first member, dividing 
by 2, and integrating the last term we have 

I sec' X dx = i [sec x tan x + log (sec x + tan x)] + C. 

Exercises 

1. I csc^ X dx. 5. j VoM-^ rfx. 

2. jsec" X dx. 6. J Vx^ — 4x + 11 (ia;. 

,. dx. 7. I , dx. 

Va^ + x2 dx. 8. I \/x2 — 9 dx. 

t/3 

109. Wallis' Formulas. Formulas will now be derived which 
make it possible to write down at once the values of the definite 
integrals: 

I sin" d de, 



2 

COS" 6 dd, 
and 



f 



I sm« 9 cos" 6 dd, 



where m and n are positive integers greater than 1. 

J sin" Odd = \ sin"-i sin d dd. 
Jo 



§109] INTEGRATION 

Integration by parts gives 



205 






sin" ddd = — sin»-^ d cos B 



a 



+ (n - 1 ) I sin"-2 e cos2 B dB 
2 0(1 - sin2 B) dB 



= (n — 1) I sin""' 

= (n - 1) I sin»-2 BdB - (n - 1) \ sin» 
Jo Jo 

g the last term and dividing by n 

J sin" BdB= ^ ~ I sin"-^ B 
« Jo 



rfe. 



On transposing the last term and dividing by n we obtain 



dB. 



This equation can be regarded as a reduction formula for 
expressing 



r 



sin" B dB 



in terms of an integral in which sin B occurs with its exponent 
diminished by 2. Applying this formula successively we obtain 



I sin" BdB = 1 I ' sin"-" B dB 

Jo n n- 2X 



n — In — 3n — 5 
n 71 — 2 n — 4 

(n-l)(n-3) 



■^BdB 



n{n - 2) 
(n-l)(n-3) 



-^ ( sin BdB'iin is odd. 



t 



n{n - 2) • 
[(n-l)(n-3) 



4-2 



^r 



dB 



sin" dO 



4-2 



n(n - 2) • • • 31 
(n-l)(n-3) • • • S-Itt 
n(n - 2) • • • 4 2 2 



if n is even. 

if n is odd. 
if n is even. 



206 CALCULUS [§109 

From the fact that the integrals 



f- 



sin" X dx 
Jo 
and 

''2 



£ 



COS" X dx 



represent the areas under the curves y = sin" x and y = cos" x , 

IT 

respectively, between the limits x = and x = ^, it is clear 
from the graphs that 



I COS" xdx = I 
Jo Jo 



COS" xdx = \ sin" x dx. 



The results obtained can be expressed in the single formula 

f COS. ... ./W... - ^^^i^^^^^,-^^..) 

where a = 1 if n is odd, and a = ^ if n is even. 
In a similar way we shall evaluate 



T 

f: 



sin" d COS" 6 dd. 



I 



2 ri 

sin" d COS" d dd = | sin^-i ^ cos" d sin ^ f/0 



sin*"-^ d cos"+^ 5 






m — _ . 
+ ^ , 1 I sm'"-^ d cos"+2 d0 



n + 1 

X 

^ I ^ I sm^-2 COS" 0(1 - sm2 d) dd 

1/^2 m — 1 P 

J I sm^-^ecos"^^^ - ^^-pj I siw-dcos^dde. 



n + 



§109] INTEGRATION 207 

Transposing the last term to the left member of the equation 



b-m]£ 



sin" 6 COS" 6 do = 



sin" 6 COS" 6 dd 



m-1 C^ . 

I , I SI 

^ + 1 Jo 

m-1 r 



sin^-^acos"^^^ 



Apply this formula successively and obtain 



r 



sin" 9 cos° d dd 



(w — 1) (w — 



m — 1 m—S 



m -\- n m -{- n 



3_ p 

-2J0 



sm'"-2 0cos"ed^. 



sin"'"''^ COS" 0d^ 



{m -\- n) {m -\- n 

(m-1) (m-3) 



Z^pTV(^pcos«ed^ if 



m IS even 



(m + n) (m + n — 2) 
(m-i)(m-3)- • •i-(n-i)(n-3) 



■^ n 

•(^ + 3)J„ 



sin 6 COS" ^ dd if mis odd 



(in+n)(mH-n— 2) 
(m-i)(in-3)- 



•(n+2)(n)(n-2) 
i-(n-i)(n-3)- 



(m+n)(m+n— 2) • • • (n+2)(n)(n— 2) 
(m-i)(m— 3) • • • 2 



IT., . 

if n is 



2 2 



if n is 
odd 



and 

m is 
even. 



, w , N / , x/ ; — r if n is either even or 

(m+n)(m+n— 2) • • • (n+3)(n4-i) j •, j • jj 

' odd, and m is odd. 

The right-hand member of the last formula of this group can 
be put in a form similar to the others by multiplying numerator 
and denominator by (n — l)(n — 3) • • • 2 or 1. It becomes 

(m-i)(m-3) • • • 2 • (n-i)(n-3) • • • 2 or i 

(m+n)(m+n-2) • • • (n+3)(n+i) (n-i) (n-3) • • • 2 or i 

IT 

These formulas for | sin"* ^ cos" 6 dd can all be expressed in 
the single formula 

X 

r sin-d cos"d dd = (m-i)(m-3)--2ori(n-i)(n-3)-2ori 

Jo (m + n) (m + n- 2)---2 or I 



«• 



(2) 

IT 

where a = 1 unless m and n are both even, in which case a = „* 



208 CALCULUS [§109 

Illustration 1. By formula (1) 

_ 8-6-4-2 128 



r 
r 

B; 



sin^ Odd 

9-7-5-31 315 

Illustration 2. 

X 

cos*edd = ^!!: = ^TT. 
4-2 2 16 

Illustration 3. By formula (2), 

4.0.0 1 
sm^ X cos' X dx = = __. 

8-6-4-2 24 

Illustration 4. 



J 2 . 4-9'^-1 S 

sm^x cos^a: dx = ^ '^ = _r_. 
9-7-5-31 315 



Illustration 5. 






sin'x cos^a; dx 



_ 5-31-31 X 3t 



10-8-6-4-2 2 512 
Exercises 



cos" X dx. 



1. I sin^ e de. 7. I 
Jo Jo 

2. I cos^^ede. 8. I sin^<^d<^. 

r 

3. I cos^ede. 9. I sin^ a; COS* X dx. 



4. I sin'^ede. 10. I sin^ X cos" X dx. 
Jo Jo 

X X 

6. I COS* e do. 11. I sin< x cos^ x dx. 
Jo Jo 

x^ !L 

6. I sin«fldfl. 12. I sii 
Jo Jo 



sin' <f> cos ff> d4t. 



§110] INTEGRATION 209 

13. I sin* X cos* x dx. 16. I x^ {a^ — x*) dx. 

14. I cos^ X sin6 x dx. 17. I (a* - x^)^^^. 
Jo Jo 

15. I (a2-x2)^dx. 18. I x(a^ - x^)^dx. 
Jo Jo 

19. j a}{\ - cos e)HQ = 4a2 j sin^ | dd. 

Let 0' = 2- Then d0 = 2de' and «' = | wnen fl = tt, and fl' = 
when = 0. Hence 

a}\ (1 - coseyde = 8a2 ( sin^O'dO'. 
Jo • Jo 

WaUis' formula can now be applied. 

By transformations similar to the foregoing many integrals can be 
put into a form to which Wallis' formulas can be applied. 

20. I cos* 2ede = \ I cos* e' do'. 
Jo Jo 

r _ 

22. I x\/2ax - x* dx. 



21. I (2ax - x*)^dx. (Substitute x = 2a sin* 6.) 

I 

»2a 



fl. sixi X I b cos X 

110. Integration of f — . — ^— dx. Integrals of this 

' c sm X + d cos X 



X 



form can be reduced by the substitution, z = tan g. In making 

this substitution it is necessary to express sin x, cos x, and dx in 
terms of z. This is easily done as follows. (The student is ad- 
vised to observe the method carefully, but not to learn the results 
as he can readily obtain them whenever needed.) Since 

z = tan ^> 

X = 2 tan~* z, 
14 



210 CALCULUS [§110 

and 



dz 

dx = 2~. — j — 7, 



Further, 

and 

Then 

and 



COS;^ = 



2 X \ X V'l+2=' 

sec 2 -y/l + tan2 o 



. X ^ X X 

sin 7. = tan 7=. cos ^ 



2 2 ^"-^ 2 VlT 



„ . a; a; 2z 

sin X = 2 sin ^ cos ^ = 



2 2 1+^2 



X X 1 - z^ 

cos a; = cos^ 7^ — sin^ - = 



'■/r 



2 2 1 + z2 



dx 

Illustration 1. I ^ — r~j On making the substitution 

' ^ + 4 cos X 



X 

z = tan ^ we obtain by using the values just found for cos x and 
dx in terms of z,^ 

2dz 



! 



l-\-z' ^ 2 ' ^^ 



1 + 4, , _3 



^/r 



1 -z^ J 1 +2^ + 4(1 -z^) 

1 +2^ 

dz 



-/^ 



322 

2 C \/3dz 



■J- 



VsJ 3z2 - 5 

2 ,„gV^i^- + c 



2\/3 \/5 \/3z + VS 

V3 tan I + VS 

= A \/l5 log ^ 3 + C. 

\/3 tan 2 — \/5 

1 The student will derive these values in each problem worked in order to famil- 
iarize hinist'lf with the method. 



§111] INTEGRATION 211 

Illustration 2. | _ #:7— • Let z = tan % Then 

2dz 

dz 



f dx ^ r i+g' _ ^ r 

j5-3sinx I Qz ~ ^J 



5 + 522 - 62 

1+02 

dz 



. f dz _ , r 

V .2 _ e, + 1 - J 



_ (2-|)^ + ii 

= f • I tan-i -^ + C = ^ tan-i ^^^ + C' 

5 tan 2 ~ 3 
= § tan-i -. + C. 



Exercises 

The student will find cos x, sin x, and dx in terms of the new vari- 
able in each of the exercises. 



. r dx r 

^' J 3 + 5 cos x' ^' J 1 

i. C l^- 7 fs 

J 5 — 3 cos X J 1 

, r ^-^ . 8 ' r ^ 

J 4 — 5 sin a; * J sin a 

/ sin xdx r 

• J 2 + sin x' ^- J : 

/COS iC 1 

o I o 1^^- 10. I 

3 + 2 cos T I . 



— 3 sin X 

+ 4 sin x 
+ 2 sin X 

+ sin X 



dx 
dx. 

dx. 



x(l + cos x) 
dx 



3 sin 2x 
dx 



4 — 5 cos 2x 



111. Partial Fractions. A rational fraction is the quotient of 
two polynomials, e.g., 

ttox™ + flix"*-^ + • • + am-ix + a„ 4>{x) 



box" + bix"-^ + • • • + b„-ix + 6„ /(x) 



(1) 



1 The integrand is not in the form given in the heading of this article, but the sub- 
stitution z = tan 2 enables us to transform any expression containing only integral 

powers of sin x and cos x into a rational function of «, i.e., into a function containing 
only integral powers of z. 



212 CALCULUS [§111 

If the degree of the numerator, m, is greater than or equal to the 
degree of the denominator, n, the fraction can be transformed by 
division into the sum of a polynomial and a fraction whose nu- 
merator is of lower degree than the denominator. In this case 
the division is always to be performed before applying the 
methods of this section. 

The integration of a rational fraction cannot in general be aC' 
complished by the methods which have been given if the degree 
of the denominator is greater than 2. Illustrations will now be 
given of a process by which a rational fraction can be expressed 
as the sum of fractions whose denominators are either of the first 
or second degrees. 

Illustration 1. 

x2 + 2 



/. 



dx. 



-2x2-9x4-18 
Factoring the denominator 

X' - 2x2 - 9x + 18 = (x - 2)(x - 3)(x + 3). 
Assume 

x2 + 2 A ,_B_ C 



x3-2x2-9x + 18 X -2^x -3"^x4-3' 

where A, B and C are to be so determined that this equation shall 
be satisfied for all values of x. Clearing of fractions 

x^ + 2 = Ax2 - 9A + 5x2 + Bx - QB + Cx^ - 5Cx + 6C 
= {A+ B + C)x2 + (B - 5C)x - 9A - 65 + 6C. 

On equating the coefficients^ of x^, x, x", we obtain the following 
three equations for the determination oi A, B and C. 

A-hB + C = 1. 

5 - 5C = 0. 

- 9A - 6B 4- 6C = 2. 

1 In applying this process use is made of the fact that if two polynomials in x are 
identically equal, the coefBcients of like powers of x are equal. Thus, given the 
identity 

a«" + aia;"-l + • • • + ccn-i x + a„ = /Sox" + /3ii"-l + • • • +0n-i x + P„, 
then 

ao = Po 
ai = /Si 



§111] INTEGRATION 213 

From these equations 

A= -h 
5 = V. 

C = U. 



Hence 



x^ + 2 -6 11 11 



a;3 _ 2x2 _ 9a; _|_ 18 5(3. _ 2) ' 6(x - 3) ^ 30(a; + 3) 
and 



r x^+2 ^ r_rfx 

J a;3 _ 2x2 - Qx + 18 ^ ^J X- 

= - l\og{x-2) + V log (x-3)+U log (x + 3) + C. 



2 
+ V Tr-. + U r^ 



Short Method. The foregoing method of determining the 
values oi A, B, • • •, by equating coefficients of like powers of 
X, is perfectly general. However, a shorter method can sometimes 
be used. Thus in the illustration just given write the result of 
clearing of fractions in the form 

a;2 + 2 =A(x- S)ix + 3) + 5(x - 2)(a; + 3) + C(x -2)(x- 3). 

Since this relation is true for all values of x, it is true f or x = 2. 
On setting x = 2, we obtain 

6 = - 5A. 
Hence 

A = -i 
On setting x = 3^ we obtain 

11 = 6B. 
Hence 

B = V. 
On setting x = — 3, we obtain 

11 = 30(7. 
Hence 

/n* _ -Li 

Illustration 2. 

x2 + l 



/ 



(x + lKx-l)'^''- 



214 CALCULUS [§111 

Let 

x' + l _ A B C D 

(x + l){x - 1)3 ~ X + 1 "•" (x - 1)3 + (x - 1)2 "^ a; - 1* 

On clearing of fractions, 

x^ + I = Aix - ly + Bix + 1) +C{x-l){x-{-l)+D(x-iy{x+l), 

or 

x^ + 1 = Ax^ - 3Ax2 + 3Ax- A+ Bx + B + Cx^ - C 

+ Dx^ - Dx^ - Dx -j- D. 
In the first form put x = 1. Then 

5 = 1. 
In the first form put x = — 1. Then 
- 8A = 2. 
Hence 

A= -h 
Equating coefficients of x' in the second form 

A + D = 0. 
Hence 

D = - A = I 

Equating coefficients of x^ in the second form, 

-SA + C - D = I. 
Hence 

O— 14-1-4 — 2' 

Consequently 

x^-\-l ^ -1 1 1 1 

(x + l)(x - ly 4(x + 1) "*" (x - ly "^ 2(x - 1)2 ■+" 4(x - 1) 
and 

J (X + 1)(X-1)3'''' ^Ja^ + l+J (.r_l)3 

= _ ^ log (X + 1 ) - 2^^^, - 2(^ + i ^ 



§1111 INTEGRATION 215 

Illustration 3. 



/i 



3.r2 - 2a; + 2 

ax. 



{x - l)(a:2-4x + 13) 

Let 

3x2 -2a; + 2 ^ 5.r + C 



(x -l)(a:2-4x + 13) x-1 ' x^ - 4a; + 13 
Clearing of fractions, 

3x2 - 2x + 2 = A{x^ - 4x + 13) + Bx(x - 1) + C(x - 1), 
or 

3x2 - 2x + 2 = Ax^ - 4Ax + 13A + 5x2 ^ Bx + Cx - C. 

In the first form put x = 1. We obtain 
3 = lOA. 
Hence 

A = A. 

Equating the constant terms in the second form, 

13A - C = 2. 
Hence 

18 - C = 2 

and 

n — i-9 

O — 1 u- 

Equating the coefficients of x^ in the second form, 

A+ B = 3. 
Hence 

5 = 3 — "i = To . 
Consequently 

C 3x2-2x + 2 . 3 r rf^ , 1 r 27x + 19 

J(x-l)(x2-4x + 13)^^ = '^"J^^ + "^"Jx-2-4x + 13''^ 

= A log (x-1) + ,ij ^ ._4^^i3 + i3 J (^^ 



dx 



-2)2 + 9 

X — 2 
= -h log (x-1) + 2-& log (x2-4x+13) + H tan-i -3— + C 

Illustration 4. 

2x dx 
(r+x)(l+x2)5 



Ji 



216 CALCULUS [§111 

Let 2a; ^ Bx -\- C Dx + E 

(1 + x)(l + x^y ~ 1 + a; "•" (1 + x^y "^ (1 + x-")' 

In Illustrations 1 to 4 a fraction was broken up into "partial 
fractions." The denominators were the factors of the denomina- 
tor of the given fraction. In Illustrations 1 and 2 the factors were 
all real linear factors, while in Illustrations 3 and 4 there were also 
factors of the second degree which could not be factored into two 
real linear factors. The method of procedure will be further indi- 
cated by the following examples. They will be grouped under the 
numbers I, II, III, and IV, corresponding to Illustrations 1, 2, 3, 
and 4. 

I. Factors of denominator linear, none repeated. 

, . x' + 5 ^ g C 

^"^ {x - l)(x + l)(a; -3) a;-l"'"x + la;-3 

., . x^ -I- 2a: + 7 A_ B 

^' (a; + 4) (2a; 4- 3) (x- 2) (3a; + 1) a; + 4 "*" 

+ ^ 




a; - 2 ' 3a; + 1 
II. Factors of denominator linear, some repeated. 

a;^ + 2a; + 5 ^_ . ^_ . C' 



(x-2)2(a;-3)'(x + l) (a; - 2)^ ' x-2 ' {x-ZY 

D E 



(a; -3)2 ' a; -3 ' x + 1 



... x» + 4a; - 2 A B 



(2x -f l)8(a; + 3)(a; - 4)^ (2a; + l)^ ^ (2a; + 1)^ 



2x + 1 ' x + S ' (x - 4)2 ' a; - 4 

III. Denominator contains factors of second degree, none 
repeated. 

x2 + 7x + 3 _ Ax + B C 

^^' (x2 + 4) (a; - 2) ~ x* + 4 + x - 2* 

„ . x» - 3x + 5 Ax-\-B Cx-^D 

"' /•-V.2 _i_ o^/•^2 _ /i^ j^ '7\/'^ 1 o^ "~ ™2 I o I 



(x2 + 2)(x2 - 4x + 7)(x + 3) x2 + 2 ^ x2 - 4x + 7 

+ _^. 
X -\- 3 

x2 + 2x - 5 _ Ax + B C__ D 

^^^ (x2 + 7)(x-2)2 x2 + 7 "^ (x-2)2 + x -2' 



§111] INTEGRATION 217 

IV. Denominator contains factors of second degree, some 
repeated. 

a;3 + 2x2 + 5 Ax + B 



(a) 



(x2 + 2x + 10)2(x2 + 3)(a; + 2) (x^ + 2x + 10) ^ 

Cx + D Ex-^F G 



a;2 + 2x + 10 ' x^ + S ' x + 2 
Exercises 



X3 + X - 10 '^*' *• J (i+1 

r (x-4)^x r 

J X' - 6x« + 9x °' J 

r x« + x^ + 7x + 1 



dx 

KxM-T)' 



(3 + 4x - x") dx 
(x- lKx''-2x + 5)' 

5x2 _|_ 13a; _ 7 



(x+4)(2x+l)2 



dx. 






dx 

I* 

+ x3+3 



— 9x 



dx. (Divide numerator by denominator.) 



CHAPTER XIII 

APPLICATIONS OF THE PROCESS OF INTEGRATION. 
IMPROPER INTEGRALS 

112. In this section a brief summary and review of the appHca- 
tions of the process of integration will be given. 

1. Area under a Plane Curve: Rectangular Coordinates. 

nb 
A = I f{x)dx. 



See §64, and Fig. 46. 

2. Area: Polar Coordinates. 



-'£ 



See §99, and Fig. 74. 

3. Length of Arc of a Plane Curve: Rectangular Coordinates. 

dx 



i:>F(i)'*- 



See §69, and Fig. 49. 

4. Length of Arc: Polar Coordinates. 






See §98, and Fig. 72. 

5. Volume of a Solid of Revolution. 

»b 
F = I Try"^ dx 



See §68, and Fig. 49. 



218 



§112] IMPROPER INTEGRALS 219 

6. Surface of a Solid of Revolution. 



'x = b 

ds 



= 2ir I y 

Jx =a 

y ds. 



See §70, and Fig. 49. 

7. Water Pressure on a Vertical Surface. 

P = k\ uz du, 

Ja 

where z denotes the width of the surface at depth u and k = 62.5 
pounds per cubic foot if u and z are expressed in feet. See §72, 
and Fig. 50. 

8. Work Done by a Variable Force. See §67. 

Exercises 

1. Find the area in^the first quadrant between the circle x^ + y'^ = a* 
and the coordinate axes. 

The definite integral which occurs in the solution of this problem is 
of very frequent occurrence. See Illustration 3, §105. 

2. Find the area bounded by the lemniscate, p^ = a^ cos 20. 

3. Find the length of one quadrant of the circle x^ + 2/^ = o*, or 
X = a cos 0, y = a sin 0. 

4. Find the length of p = 10 cos 0. 

6. Find the volume of a sphere of radius a. 

6. A solid is generated by a variable square moving with its center 
on, and with its plane perpendicular to, a straight line. The side of 
this square varies as the distance, x, of its center from a fixed point 
on the line, and is equal to 2 when a; = 3. Find the volume generated 
by the square when its center moves from a; = 2 to x = 7. 

7. Find the area of the surface of a sphere of radius a. 

8. The unstretched length of a spring is 25 inches. Find the 
work done in stretching it from a length of 27 inches to a length oi 



220 CALCULUS (§112 

29 inches, if a force of 400 pounds is necessary to stretch it to a 
length of 26 inches. 

9. A trough 3 feet deep and 2 feet wide at the top has a parabolic 
cross section. Find the pressure on one end when the trough is 
filled with water. 

3 3 2 

10. Find the length of the curve x* +2/' =a^, orx = ocos' 6, 
y = a sin' B. 

11. Show that the work done by the pressure of a gas in expanding 
from a volume i^i to a volume V2 is given by 

I p dv. 

Jn 

where p is the pressure per unit area. 

Hint. Take a cylinder closed by a piston of area A forced out 
a distance Ax by the expanding gas. Denote by At/; the work done 
by the gas in expanding from a volume r to a volume v -\- Av, 
Then, 

I] 

p dv. 



-f 



12. Find the area of one quadrant of the ellipse x = o cos 6, 
y = b sin 6. 

13. Find the area of one loop of the curve p = a cos 26. 

14. Find the length of the cardioid, p = a(l — cos 6). 

15. Find the volume of the ellipsoid of revolution generated by 

revolving the ellipse "i + j^i ~ ^ about the X-axis; about the F-axis. 

16. A volume is generated by a variable equilateral triangle moving 
with its plane perpendicular to the JC-axis. Find the volume of the 
solid between the planes x = and x = 2, if a side of the triangle is 
equal to 2x^. 

17. Find the area of the surface generated by revolving about 
the X-axis the portion of the arc of the catenary 



a 
^=2 



+ e 



■■] 



between (0, a) and (xi, yi). 

18. Find the area under one arch of the cycloid x = a {d — sin d), 
y = a(l — cos 9). 



§112] IMPROPER INTEGRALS 221 

19. Find the length of that portion of 9y^ = x' above the Z-axis 
between x = and x = 3. 

20. Find the volume generated by revolving the portion of the 
catenary 



y 



« r - . --1 

2 e" +e « 



between x = and x = b about the X-axis; about the F-axis. 

21. Find the volume generated by revolving the hypocyoloid 

2 3 1 

x^ + y^ = o*, or X = a cos^ 0, y = asm^ 0, about the X-axis. 

22. Find the area included between the parabolas 47/^ = 25x and 
5x^ = \oy. 

23. Find the area between the X-axis, the curve y = x'' — 4x + 9, 
and the ordinates x = 1 and x = 7. 

24. Find the area between the curve y = sin x, the X-axis, and 
X = and x = w. 

26. If a gas is expanding in accordance with Boyle's law, -pv = C, 
find the work done in expanding from a volume vi to a volume V2. 
Represent the work graphically by an area. 

26. Find the work done if the gas is expanding in accordance with 
the adiabatic law, py* = C. 

Hint. From the result of Exercise 11, 






C 



k 



A-k _ „.l-ft 



Vi^-'). 



Now, 

C == PiVi* = ^2^2*. 

Hence 

W = j-iri; (P2i'2 - PlVl). 

Represent the work graphically by an area. Use the same scale as 
in Exercise 25. 

x^ V^ 

27. Find the area of one quadrant of the ellipse Ta + g" ~ ■'■• ^^ 

Exercise 1. 

28. Find the length of p = e"* from ^ = to ^ = 27r. 

29. Find the length of p = e""^ from ^ = to (? = - =" , if a is 
assumed positive. 

30. Find the area bounded by the cardioid p = a(l + cosO). 

31. Find the area bounded by p = 10 sin d. 

32. Find the area bounded by the hypocycloid x = o cos^ 9, 
y = a sin^ d. 



222 CALCULUS [§112 

33. Find the area between y^ = Ax and y^ = 8x — x'^. 

34. Find the work done by a gas in expanding isothermally from 
an initial volume of 2 cubic feet and pressure of 7000 pounds per square 
foot to a volume of 4 cubic feet. 

36. Find the work done if the gas expands adiabatically. Take 
k = \h the value for steam. (See Exercise 26.) 

36. Find the pressure on a trapezoidal gate closing a channel con- 
taining water, the upper and lower bases of the wet surface being 
25 feet and 18 feet, respectively, and the distance between them being 
10 feet. 

37, Find the area between the catenary 



|[^ea+e aj, 



the X-axis, and the ordinates x = and x = a. 

38. Find the length of p = ad from = to fl = 27r. 

39. Set up the integral representing the length of one quadrant 
of the ellipse x = a cos 0, y = b sin 0. 

40. Find the volume generated by a circle of variable radius mov- 
ing with its plane perpendicular to the X-axis, between the planes 
X = 2 and x = 8. The radius is proportional to x^ and is equal to 54 
when X = 3. 

41. Find the volume generated by revolving one arch of the cycloid 
X = a{d — sin d), y = a(l — cos 6) about the X-axis; about the 
tangent at the vertex. 

42. Find the area of the surface generated by revolving a quadrant 
of a circle about a tangent at one extremity. 

43. If the density of a right circular C3dinder varies as the distance 
from one base, find the mass of the cylinder. if the altitude is h and the 
radius of the base is r. 

44. The force required to stretch a bar by an amount s is given by 

„ Eos 

where E is the modulus of elasticity of the material of the bar, o is the 
area of the cross section, and L is the original length. Find the work 
that is done in stretching a bar whose unstretched length is 400 inches 
to a length of 401 inches, if £ = 30,000,000 pounds per square inch and 
o = 1.5 square inches. 

46. Find the area of one loop of p = 10 sin Zd. 



§113] IMPROPER INTEGRALS 223 

46. Find the length of 

y = ^\ 



|[^e- +e~aj 



from (0, a) to {xi, y,). 

47. Find the length of one arch of the cycloid x = a{d - sin. e) , 
y = a{l — cos e). 

48. Find the volume of the anchor ring generated by revolving the 
circle x^ + (y - by = a^ about the X-axis, a being less than b. 

a 

49. Find the area of the small loop of p = a sin^ ^ * 

o 

60. Find the work done in pumping the water out of a cistern 20 feet 
deep, in which the water stands 8 feet deep, if the cistern is a parabo- 
loid of revolution and the diameter at the surface of the earth is 8 feet. 

51. Find the volume included between two equal right circular 
cylinders, radius a, whose axes intersect at right angles. 

62. Find the area of the surface generated by revolving one arch of 
the cycloid x = a{e — sm 6), y = a{l — cos 9), about the X-axis; 
about a tangent at the vertex. 

63. Find the area bounded by p =3+2 cos 6. 

64. Find the area bounded by the small loop of p =2 + 3 cos B. 
66. Find the area of the surface generated by revolving the cardioid 

p = a(l + cos 6) about the polar axis. 

56. Find the volume bounded by the surface of Exercise 55. 

113. Improper Integrals. Since — , becomes infinite at 

■s/x — 1 

X = 1, the definite integral 

r 1 

dx 



: 



■s/x 



must not be evaluated by the usual process. For, the assumption 
has been made that in the integral 






fix) dx 



fix) is a continuous finite function at x = a and x = 6 as well as 
at all intermediate points, and the evaluation of this integral was 
based on the area under the curve y = fix). In this case 



224 CALCULUS [§113 

becomes infinite at the lower limit. The area under the curve 

1 



y 



Vx^^ 



between the ordinates x = 1 and x = 7 has no meaning. In fact 
the integral in question has no meaning in accordance with the 
definition of a definite integral already given. A new definition 
is necessary. We define 

•^ 1 

dx 



as 



r 



im / 



dx, 

+, Vx - 1 

where r/ is a positive number, if this limit exists. Otherwise the 
integral has no meaning. Now, 

1™ f -yL^ dx - >s, (2Vi^i) r 

Ji+, y/x-\ I14-, 

= I'i? (2 V6 - 2v^) = 2V6. 
Since the limit exists we say that 



J 



7 1 

■ dx = 2\/6. 



Graphically this means the limit as 17 approaches zero of the area 

under the curve y = , between the ordinates x = 1 + ?; 

V a; — 1 

and X = 7, exists and is equal to 2\/6. 

Exercise 1, Show that 



i 



dx 



], {X - 1)' 
exists if < w < \. 

On the other hand, when n = 1, 



im r 1 , lim , / ^sV 

Ji+v X I ^ 1^ 



1+17 



§113] IMPROPER INTEGRALS 225 

This limit does not exist and consequently we say that 

1 



/ 



x-1 



dx 



has no meaning or does not exist. 

Graphically this means that the area under the curve 



y = 



1 

X- 1 



between the ordinates x = 1 + t] and x = 7 increases without 
limit as rj approaches zero. 
Exercise 2. Show that 



S. 



dx 



X (^ - 1)" 

does not exist if n ^ 1. (Note that the case n = 1 has just been 
considered.) If n<0 no question as to the meaning of the 
integral can arise. Why? 

A definite integral in which the function to be integrated 
becomes infinite at the upper limit is treated in the same way. 
Thus 



I 



dx 



is defined as 

dx 



lim p"" 



vn=- 



where ?; is a positive number, if tliis limit exists. 
Exercise 3. Show that 



i 



dx 



(1-^)" 

has a meaning in accordance with this definition if < n < 1, and 
that it has no meaning if n > 1. If » < no question can arise as 
to the meaning of the integral. 

It is easy to see how to proceed in case the function under the 

15 



226 CALCULUS [§114 

integral sign becomes infinite at a point within the interval of 
integration. Thus 

dx 

;;» where n is a positive integer, 



I 



is defined as 

limrr-" dx__ r dx 

where r; is a positive number, if this limit exists. If not, the 
integral has no meaning. Tf n < no limit process is necessary. 

Exercises 
Evaluate the following integrals if they have a meaning: 

dx 



3x-4 
dx 

2 

I 3 

dx 

2)^ 



* Jo ^'' ■ Jo VoT^' • J_, X 

Jo "^ X V^^=^^ Jo (X- 

4 f^^ dx g p dx 

12. Find the area between the curve y' = ^ — — — > its asymptote and 

the X-axis. 

114. Improper Integrals : Infinite Limits. In §113, the interval 
of integration was finite. In other words neither of the limits oi 
the integral 





fmdx 


was infinite. 




The integral 






p dx 
Jo ^' + «^ 


will be defined as 






lim r dx 

6= CO 1 3-2 1 ^ 



§114] IMPROPER INTEGRALS 227 

if this limit exists. Now 



lim 

b = 



lim 1 , _,b 1 TT 

0-" a a a 2 



im r d^ ^ ^^"^-tan-i- 
6™ I 2 I — 2 represents graphically the limit of the area under 

the curve y = — ;rn — ; between the ordinatcs x = and x = h 
" x^ + a^ 

as & increases indefinitely. 

Consider 

dx 

X ' 

lim / dx lim lim , , 



inci 

I 



X 



But log 6 increases without limit as h increases witnout limit. 
Hence | — has no meaning. 



Exercises 
Evaluate the following integrals if they have a meaning: 

2. I e-'dx. 4. I x'^e-'dx. 

Jo Jo 

6. Find the area between the witch, ij = VXT^' ^^^ ^^® ^^^^ °^ ^* 



CHAPTER XIV 



SOLID GEOMETRY 



115. Coordinate Axes. Coordinate Planes. Just as the posi- 
tion of a point in a plane is given by two coordinates, for example 
by its perpendicular distances from two mutually perpendicular 
coordinate axes, the position of a point in space is given by three 
coordinates, for example by its perpendicular distances from three 
mutually perpendicular planes of reference, called the coordinate 
planes. Let the three coordinate planes be those represented in 
Fig. 75, viz., XOY, called the ZF-plane, YOZ, called the FZ-plane, 
and ZOX, called the ZX-plane. Jhen the position of the point P 
whose perpendicular distances from the YZ-, ZX-, and XF-planes 



/ 





Fig. 75. 



Fig. 76. 



are 2, 3, and 1, respectively, is represented by the coordinates 2, 3, 
and 1. The lines of intersection of the planes of reference are 
called the axes. Thus X'OX, Y'OY, and Z'OZ, Fig. 76, are called 
the axes of x, y, and z, respectively. The coordinates of a point 
P measured parallel to these axes are known as its x, y, and z 
coordinates, respectively. Thus for the particular point P 
of Fig. lb,x = 2,y = 3, and z = 1. More briefly we say that the 
point P is the point (2, 3, 1). In general, (x, y, z) is a point whose 
coordinates are x, y, and z. If these coordinates are given the 

228 



§115] SOLID GEOMETRY 229 

position of the point is determined, and if a point is given these 
coordinates are determined. 

The relation between a function of a single independent variable 
and its argument can be represented in a plane by a curve, the 
ordinates of which represent the values of the function correspond- 
ing to the respective values of the abscissas. Thus, y = j{x) is 
represented by a curve. To an abscissa representing a given value 
of the argument there correspond one or more points on the 
curve whose ordinates represent the values of the function. In 
like manner a function of two independent variables x and y can 
be represented in space. Choose the system of coordinate planes 
of Fig. 75. Assign values to each of the independent variables 
X and y. These values fix a point in the XF-plane. At this 
point erect a perpendicular to the XF-plane, whose length z repre- 
sents the value of the function corresponding to the given values of 
the arguments. Thus a point P is determined. And for all values 
of X and y in a given region of the XF-plane there will, in general, 
correspond points in space. The locus of these points is a surface. 
The surface represents the relation between the function and its 
two independent arguments just as a curve represents the relation 
between a function and its single argument. 

Thus if 2 = ± V25 -x^ - y^ = /(x,y), ± \/l2 are the 
values of the function corresponding to the values x = 2 and 
2/ = 3. Then th e points (2, 3, 2 \/3) and (2, 3, - 2V3) lie on the 

surface 2 = ± V^25 -x^- y\ If x = -3and?/ = 1, 2 = ± -s/l5- 
The corresponding points on the surface are (— 3, 1, -v/lS) 
and (-3, 1, - \/l5)- 

The coordinate planes divide space into eight octants. Those 
above the XF-plane are numbered as shown in Fig. 76. The oc- 
tant immediately below the first is the fifth, that below the second 
is the sixth, and so on. The points (2, 3, 2\/3) and (2, 3, - 2\/3) 
lie in the first and fifth octants, respectively. The points 
(-3, 1, \/l5) and ( — 3, 1, — \/l5) lie in the second and sixth 
octants, respectively. 

The locus of points satisfying the equation 

z = ± V25 - x^ -tj^ (1) 

iS a sphere of radius 5. For, this equation can be written in the 



230 



CALCULUS 



[§117 



form x^ + 2/^ + 2^ = 25, which states that for any point P on the 
surface (1), OP = Vx"^ + y^ + z^ = 5. The left member is the 
square of the distance, OP, of the point P (x, y, z), from 0, since 
OP is the diagonal of a rectangular parallelopiped whose edges 
are x, y, and z. If then the coordinates of P satisfy (1), this point 
is at a distance 5 from the origin. It lies on the sphere, of radius 5, 
whose center is at the origin. 

116. The Distance between Two Points. The student will 
show that the distance d between the two points (xi, yi, Zi) and 

(X2, t/2, Zz) is 

d = V(x, - xi^ + (i/2 - yir + (22 - zi)K (1) 

See Fig. 77. If the point (xi, yi, Zi) is the origin, (0, 0, 0), the 



expression for d becomes 
P = 



VX2^ + 2/2^ + 22= 



(2) 





Fig. 77. 



Fig. 78. 



Exercises 

Find the distance between the following points : 

1. (1, 2, 3) and (3, 5, 7). 

2. (1, -2, 5) and (3, -2, -1). 

3. (0, -3, 2) and (0, 0, 0). 

4. (0, 0, 3) and (0, 2, 6). 
6. (0, 0,-5 and (2, 0, 6). 

6. (-3, 2, -1) and (0,0, 0). 

117. Direction Cosines of a Line. Let OL, Fig. 78, be any line 
passing through the origin. Let a, jS, and y be, respectively, the 
angles, less than 180°, between this Hne and the positive direc- 
tions of the X-, Y-, and ^-axes. These angles are called the 



§118] 



SOLID GEOMETRY 



231 



direction angles of the line, and their cosines are called the direction 
cosines of the line. Let P, whose coordinates are x, y, and z, be 
any point on the line. Let OP = p. Then 

X = p cos a, 

y = p cos /3, 
and 

z = p cos y. 
Squaring and adding the above equations we obtain 

2-2 _|. y2 _^ 2^ = p2(cos^ a + cos^ /3 + cos^ 7). 
Since 

3.2 _j_ j^2 ^ 2^ = p2, 

cos^a + cos^/3 + cos^Y = L (1) 

The direction cosines of any line are defined as the direction 
cosines of a parallel line passing through the origin. Then, the 
sum of the squares of the direction cosines of any line is equal to 
unity. 

Exercises 

Find the direction cosines of the lines passing through each of the 
following pairs of points. 

1. (0, 0, 0) and (1, 1, 1). 

2. (0, 0, 0) and (2, -3, 4). 

3. (0, 0, 0) and (-1, 2, -3). 

4. (1, 2, 3) and (5, 6, 7). 
6. (-2, 3, -1) and (-3, -4, 3). 

118. Angle between Two 
Lines. Let AB and CD, Fig. 
79, be two lines, and let their 
direction cosines be cos aj, cos /3i, 
cos 7i, and cos 0:2, cos ^2, cos 72, 
respectively. Denote the angle 

between the lines by d. Let CH, HK, and KD be the edges of 
the parallelopiped formed by passing planes through C and D 
parallel to the coordinate planes. The projection of CD on AB 
is clearly equal to the sum of the projections of CH. HK, and 
KD on AB. 
Hence 

CD cos d = CH cos ai + HK cos /3i + KD cos 71. 




232 



CALCULUS 



[§119 



Now 



and 



CH = CD cos a2, 

HK = CD cos /32, 



KD = CD cos 72. 
Consequently 

CD cos = CZ)(co8 ai cos ccz + cos /3i cos /32 + cos 71 cos 72). 
Hence 

cos 6 = cos ai cos 0:2 + cos |8i cos 182 + cos 71 cos 72. (1) 

Exercises 

Find the cosine of the angle between the lines determined by the 
points of Exercises 1 and 2; 2 and 3; 3 and 4, of the preceding 
section. 

119. The Normal Form of the Equation of a Plane. — ^Let ABC, 

Fig. 80, be a plane. Let ON, 
the normal from 0, meet it in 
N. Let the length of ON be p 
and let its direction angles be a, 
j3, and 7. If p, a, /3, and 7 are 
given the plane is determined. 

We seek to find the equation 
of the plane. Let P, with co- 
ordinates X, y, and z, be any 
point in the plane. The sum 
of the projections of OH = x, HK = y, KP = z, and PN upon 
ON is ON = p. 

The projection of OH on ON is x cos a. 
The projection of HK on ON is y cos j8. 
The projection of KP on ON is z cos 7. 
The projection of PN on ON is 0. 
Hence 

. X cos a + y cos ^ + z cos 7 = p. (1) 

If P does not lie in the plane ABC, the projection of PN on OiV 
is not zero, and the coordinates of P do not satisfy (1). Hence the 
locus of a point satisfying (1) is a plane. Equation (1) is the 
normal form of the equation of the plane, p is taken to be 




§120] SOLID GEOMETRY 233 

positive. The algebraic signs of cos a, cos/3, and cosy are de- 
termined by the octant into which ON extends. 

Illustration 1. Find the equation of a plane for which p = 2, 
a = 60°, 13 = 45°. 

cos a = ^, 

cos /3 = — -^. 
V2 
Then by (1), §117, 

cos- 7 = 1 — J — §• 
Hence 

cos 7 = ± i. 

The equation of the plane is 

2 + V2 - 2 

There are thus two planes satisfying the conditions of the problem, 
one forming with the coordinate planes a tetrahedron in the first 
octant, the other a tetrahedron in the fifth octant. 

Exercises 

1. Find the equation of a plane if a = 60°, /3 = 135°, p = 2, and 
if the normal ON extends into the eighth octant. 

2. If a = 120°, /3 = 60°, p = 5 and if the normal ON extends into 
the sixth octant. 

120. The Equation Ax + By + Cz = D. The general equation 
of the first degree in x, y, and z is 

Ax + By ^ Cz = D, (1) 

where A, B, C, and D are real constants. D may be considered 
positive. For, if the constant term in the second member of an 
equation of the form (1) is not positive it can be made so by 
dividing through by —1. 



.(2) 



Divide (1) by VA^ + B^ 
A ^ , 


+ C2 
B 


and obtain 


D 




VA^ -\- B^ + C^ 


Va^ 

1 


+ B^ 
C 


+ C2^ 






^VI 


' + B- 


2 + C2 \/A^ 


+ B^ 


+ C^' 



234 CALCULUS [§121 

The coefficient of x is either equal to or less than unity in numerical 
value. It can then be considered as the cosine of some angle, say 
a. Similarly the coefficient of y may be considered as the cosine 
of some angle /3, and that of z as the cosine of some angle 7. 
Further the sum of the squares of these coefficients is equal to L 
Hence a,^, and 7 are the direction angles of some line. Then (2) 
is in the form 

X cos a -^ y cos /3 + 2 cos 7 = p, (3) 

where 

and cos a, cos /3, and cos 7 are the coefficients of x, y, and z, respect- 
ively, in equation (2). Hence (3) is the normal form of the equa- 
tion of a plane. Equation (1) is the general equation of the first 
degree in the variables x, y, and z. Therefore every equation of 
the first degree in x, y, and z represents a plane. 

Illustration 1. Put 3x — 2y — z = Q in the normal form. 
Divide by VZmTbM^ = \/9 + 4 + 1 = \/U and obtain 
3j 2y z__ _ 6 

Vii \/l4 Vii ~ Vii 

The plane is . — units distant from the origin, and forms, with the 
Vl4 

coordinate planes, a tetrahedron in the eighth octant. 

Exercises 

Transform each of the following equations to the normal form, find 
the distance of each plane from the origin, and state in which octant 
it forms a tetrahedron with the coordinate planes. 

1. Sx - 2y - z = 1. 6. X + 2?/ = 6. 

2. X + y + z = —1. 7. x — 2 = 4. 

3. X - 3r/ + 2z = 3. 8. x = 2. 

4. x - 2j/ + 3z + 2 = 0. 9. X = -1. 
6. 2x - y - z - 1 = 0. 10. z = y. 

121. Intercept Form of the Equation of a Plane. We seek the 
equation of a plane whose intercepts on the X-, Y-. and Z-axes 
are a, b, and c, respectively. 



§121] SOLID GEOMETRY 235 

The general equation of a plane is 

Ax + By -{- Cz = D. (1) 

The constants are to be so determined that the plane will pass 
through the points (a, 0, 0), (0, b, 0) and (0, 0, c). 

On substituting the coordinates (a, 0, 0), in (1), we obtain 

Aa = D, 
or 

a 

Similarly, since (1) passes through (0, b, 0), 

Bb = D, 
or 

-?• 

And, since it passes through (0, 0, c), 

Cc = D, 
or 

C 

With these values oi A, B, and C, (1) becomes 
a c 



or 

a ' b ' c 



+ f + - = 1- (2) 



Equation (2) is known as the intercept form of the equation of a 
plane. 

Illustration. Transform the equation 3x — 2?/ — 52 = 4 to the 
intercept form. Divide by 4 and obtain 

-+-^ + -^=1 

3 ■^ 5 

The intercepts on the X-, F-, and Z-axes are i, —2, and — |, 
respectively. 



236 CALCULUS [§123 

Exercises 
Transform each of the following equations to the intercept form : 

1. X +y +z = 3. 4. 2a; + 7y - 3z = 1 

2. 2x - 3y + 42! = 7. 5. x - y + Sz = -\. 

3. 2x + y - z + 2 = 0. 6. t/ - 2x - 3z = 5. 

122. The Angle between Two Planes. The angle between two 
planes is the angle between the normals drawn to them from the 
origin. The cosine of the angle between the normals can be found 
by formula (1) §118, in which ai, /3i, 71 and 0:2, ^2, 72 are the direc- 
tion angles of the normals. 

Illustration. Find the angle between the planes 

x + y + z = l (1) 

and 2x + y + 2z = S. (2) 

Transform these equations to the normal form and obtain 

Vs Vs V^ Vs 

and 

2a; V 22 

3+1+3=1. (4) 

The direction cosines of the normals to the first and second 

planes are —t=, —j=, — ^, and f , \, f , respectively. Then, if Q is 

the angle between the normals, formula (1), §118, gives 

fl 2,1,2 5 

cos Q = - — 7= + - — 7^ + 



3\/3 3\/3 3\/3 3\/3 
From which Q = 74.5°. 

• Exercises 
Find the angle between the following pairs of planes : 

1. X - 3j/ + 2z = 6 and x - 2?/ + 2 = 1. 

2. X - 2?/ + 3z = 2 and 2x + y - 2z = 3. 

123. Parallel and Perpendicular Planes. If two planes are 
parallel = and cos Q — \. If they are perpendicular Q = 90° 
and cos 6 = 0. 
Let 

Arx-^ B,y + C,z = Di (5) 

and 

AiX + B^y + C2Z = D2 (6) 



§124] SOLID GEOMETRY 237 

be the equations of two planes. After writing these equations in 
the normal form it is found that 

AiA2 + BiB2-\-CiC2 
cos d = , (7) 

If AxA^ + B,B^ + C1C2 = 0, (8) 

cos B = Q and the planes (5) and (6) are perpendicular. 

If the planes (5) and (6) are parallel, the corresponding coeffi- 
cients must be equal or proportional. For then and only then will 
their normals be parallel. 

Exercises 

From the following equations pick out pairs of parallel planes and 
pairs of perpendicular planes. 

1. a; + J/ + z = 6. 

2. X — y - z = 2. 

3. 2x + 2y + 2z = 7. 

4. 3x - 27/ - z = 8. 
6. 2x - 3i/ + z = 1. 

124. The Distance of a Point from a Plane. Let {xi, iji, Zi) 
be any point and let 

Ax + By + Cz = D 
be the equation of a plane. We shall find the distance of the 
point from the plane. 
Now 

Ax + By + Cz = K, 
where K is any constant, is the equation of a plane parallel to the 
given plane. (See §123.) Let us choose K so that this plane shall 
pass through the given point (xi, ?/i, Zi). To do this substitute 
the coordinates of the point in the equation and solve for K. 
This gives 

K = Axi + Byi + Czi. 
Placing the equation of each plane in the normal form we have 

Ax-hBy + Cz D 



and 


R 

Ax + By + Cz K 
R R ~ 


R 

Axi + Byi + Czi 
R 


where R 


= VA^ + B^ + C. 





238 



CALCULUS 



[§125 



The given plane is -h units distant from the origin, and the plane 

through the point (xi, yi, Zi) is p — units distant 

from the origin. Then the distance, d, between the two planes, 
and hence the distance of the given point from the given plane, 
is equal to the difference of these two distances, or 
d = Axi + Byi + Czi - D 

\/A24^B2"+C2 

Illustration. Find the distance of the point (1, 2, —1) from the 
plane 3x — y + z-{-7 = 0. 

Axi + Byi -\-Czi-D _ 31 - 1-2 + !•( - 1) +7 ^ 7 

Vn 



d = 



VA^ + B^ + C^ 



V32+(-l)2 + P 



Exercises 



In each of the following find the distance of the given point from 
the given plane: 

1. (3,1,-2); 3x + y - 2z - 6 = 0. 

2. (-1,2,-3); x-y -2z + l =0. 

3. (0, 2,-3); 2x + 3y - 5z - 10 = 0. 

125. Symmetrical Form of the Equations of a Line. Let PPi, 

Fig. 81, be a line passing through 
the given point Pi (xi, yi, zi), 
and having the direction cosines 
cos a, cos /3, cos y. In order to 
find the equations of the line, 
let P (x, y, z), be any point on 
the line and denote the distance 
PPi by d. Then 



Fig. 81. 
and therefore 




Xi 





X — Xi 


= d cos a, 




y - 2/1 


= d cos /3, 




Z — Zi 


= d cos 7, 


y - 2/1 


Z - Zi 




cos j3 


cos 7 





(1) 

COS a — 

These equations are known as the symmetric equations of the 
straight line. 



§125] SOLID GEOMETRY 239 

Frequently a straight line is represented by the equations 
of two planes of which it is the intersection. 
Illustration 1. 

3x-y + l=0, (2) 

5x -z = S. (3) 

From these equations the symmetrical form of the equations can 
readily be obtained. From (2) and (3) we obtain 
_y-l _z+3 

"^ ~ 3 - ^5~' 
or 

x-0 y-1 z + 3 ... 

nr = ~3~ = ~5- (^> 

The denominators, 1, 3, and 5, of (4) are not the direction cosines 
of the line, but they are proportional to them. Upon dividing 
each by \/35, the square root of the sum of their squares, they 
become the direction cosines. Then 

X -0 _ y-1 _ 2 + 3 
1 ~ 3 ~ 5 



-s/35 \/35 VS5 
is the symmetrical form of the equations of the line. 

The line therefore passes through the point (0, 1,-3) and has 
the direction cosines given by the denominators in the preceding 
equations. 

Illustration 2. Consider the line which is the intersection of 
the planes 

13a: + 52/ - 42 = 40, 
-13x + lOy -2z = 23. 
On eliminating x we obtain 

5y - 2z = 21, 
and on eliminating y we obtain 

13x -2z = 19. 
From the last two equations we find 

5y - 21 13x - 19 

' = —2~ = 2 ' 

or 

x-n y-\^ z - 



A i 1 



240 



CALCULUS 



[§126 



These are the equations of a line which passes through the point 
(+3, ^6*", 0) and whose direction cosines are proportional to A, 
5, and 1. The student will find the direction cosines. 

In Illustration 1, equation (2) represents a plane parallel to the 
^-axis whose trace in the XF-plane is the line 3x — 7j -{- 1 = 0. 
Equation (3) represents a plane parallel to the F-axis whose trace 
in the ZX-plane is the line 5x — z = 3. 

In Illustration 2 the position of the two planes which intersect 
in the straight line is not so evident. By eliminating first x and 
then y, the equations of two planes passing through the same line 
are obtained, one of which is parallel to the X-axis and the other to 
the y-axis. 



Exercises 
Put the equations of the following lines in the symmetrical form : 

1. X + 2y +3z =6, 
X — y — 2 = 1. 

2. a; + y — z = 1, 
a; - 3y + 2z =6. 

Z. X - y + 2z = 0, 

X + 2y - 3z =0. 

126. Surfaces of Revolution. Let 



2/2 = 42 



(1) 




Fig. 82. 



be the equation of a curve in the 
FZ-plane, Fig. 82, and let it be 
rotated about the Z-axis. The sur- 
face generated is a surface of revolu- 
tion. Any point D on the curve de- 
scribes a circle of radius CD, equal to 
the 7/-co6rdinate of the point D. 
During the revolution the z-co6rdinate 
does not change. Let P be any posi- 
tion taken by D in the revolution. Let 
the coordinates of P be (x, y, z). 



But by (1), 



x2 -f 7/2 = icpy = {coy 

{CDY = iz, 



(2) 



§126] SOLID GEOMETRY 241 

where z is the common ^-coordinate of Z) and P. Then (2) 
becomes 

x2 + 2/2 = 42, (3) 

an equation satisfied by any point on the surface of revolution. 
We note that (3) is obtained from (1) by replacing ^/^ by x^ + y"^, 
or y by -s/x^ + y"^. 
In general, if 

/(y, 2) = (4) 

is the equation of a plane curve in the FZ-plane, the equation of 
the surface of revolution generated by revolving it about the Z- 
axis is obtained by writing -s/x^ + y"^ for y, i.e., the equation of 
the surface of revolution is 

fiVx^TV', ^) = 0- (5) 

This equation can also be regarded as the equation of the surface 
generated by revolving the curve f{x, z) = 0, lying in the XZ- 
plane, about the Z-axis. 

Similarly, /(i/, V^M^) = o (6) 

is the equation of the surface generated by revolving the plane 
curve /(y, x) = about the F-axis; and 

<A(x, V^M=^) = (7) 

is the equation of the surface generated by revolving the plane 
curve 0(.T, 2) = about the X-axis. 

Illustration 1. The equation of the surface generated by rotat- 
ing x^ -{• iy — ^y = a^ about the X-axis is 

x^ + [Vy^ + 2' - ^Y = a^ 

Exercises 

Find the equation of the surface generated by rotating: 

1. y = x^ about the F-axis. 

2. y =^ x^ — a^ about the X-axis. 

3. 6^x2 + a^yi = a^b^ about the X-axis. 

4. 6^x2 - 0^2/2 = a^b^ about the X-axis. 
6. &2x2 — a-y^ = a-b^ about the F-axis. 

6. x2 + y^ = a^ about the F-axis. 

7. x2 -f- t/2 = a2 about the X-axis. 

8. y = mx about the X-axis, 

9. J/ = mx about the F-axis. 

16 



242 ■ CALCULUS [§127 

127. Quadric Surfaces. Any equation of the second degree 
between x, y, and z, of which 

Ax^ + By^ + Cz^ + Dxy + Eyz + Fxz -\- Gx -{- Hy + Kz -h L = 

(1) 
is the general form, represents a surface which is called a quadric 
surface, or conicoid. 

By a suitable rotation and translation of the axes, the equation 
of any quadric surface can be put in one of the following forms: 
• x^ v^ z^ 

± ^' ± P ± ^ = 1' (2) 

rp2 y2 2*2 

<^ ± P ± C-' = »■ (3) 

X^ 7/2 

-,±i,= + 2c2. (4) 

The particular form assumed by the equation depends upon the 
values of the coefficients in (1). 
The quadric surface 

X^ 0/2 2j2 

^= + f= + ,T = l (5) 

is called the ellipsoid. To find the shape and properties of this 
surface, let 

X = k, (6) 

where k is any real constant. This equation represents a plane 
perpendicular to the axis of x. Equations (5) and (6) considered as 
simultaneous equations represent the curve of intersection of the 
ellipsoid with the plane. If x is eliminated between (5) and (6) 
there results 

t/2 ^2 



/ by/a^-k^ y ^ / cVa^ - k^ V 



the equation of the curve of intersection in the plane x = k. 
Equation (7) is the equation of an ellipse. The semi-axes of the 

„. &\/a2 - k^ , c\/a^ - k^ rru u * 

ellipse are —^ and —^ . These axes grow shorter 

a, a 

as A; increases in numerical value from to a. When k = ± a 



§127] SOLID GEOMETRY 243 

the elliptical section reduces to a point. When | A;| > a, the lengths 
of the axes of the ellipse become imaginary, i.e., the plane x = k, 
(,\k\ > a), does not meet the surface (5) in real points. Hence the 
surface is included between the planes z = + a. 

The above discussion shows that the surface represented by the 
equation (5) is included between the planes x = + a; is symmet- 
rical with respect to the FZ-plane; and has elliptical sections 
made by planes perpendicular to the axis of x. These sections 
grow smaller as the cutting plane is moved away from the YZ- 
plane and at a distance + a reduce to a point. 

In a similar manner, by taking y = k, and then by taking 
z = k, the student will discuss plane sections of the ellipsoid (5) 
perpendicular to the F-axis and to the Z-axis. 

a, b, and c, are called the semi-axes of the ellipsoid. 

It can be shown that any plane section of the ellipsoid is an 
ellipse. 

The surface represented by 



/2 



^^ . r _ 1 _ 1 rsi 

^2 + ^,2 c^ - 1 - w 

will now be discussed. Let z = k. Then 

S + f: = i + l' (») 

is the equation of the plane section made hy z = k. It is an 

ir . ■ aVcM^2 , bVc^ + k^ rp, 
ellipse whose semi-axes are — ' and — ^^ ■ . They 

increase in length with the numerical value of k. The axes 
have a minimum length when k = 0. The surface represented by 
equation (8) is symmetrical with respect to the XF-plane, and 
every section parallel to this plane is an ellipse. The smallest 
elliptical section is that made by the XF-plane. 
li X = k, equation (8) becomes 

P - c"^ = 1 - a^' (10) 

an hyperbola. 

If A; < a, the transverse axis of the hyperbola is parallel to the 



244 CALCULUS [§128 



F-axis. If fc > o, the transverse axis is parallel to the Z-axis. 
When k = a, equation (10) reduces to 

or 



r z'^^^ 



(^+^)(i-:-)=». 



the equation of two straight lines. 

The student will discuss the curves of intersection of the surface 
(8) with planes parallel to the XZ-plane. 

The surface is called the hyperboloid of one sheet, or of one 
nappe. 

Exercises 

The student will discuss the following surfaces and make sketches 
of them: 

a;2 y2 z2 

1. —2~ hi 2 ~ ^» ^^^ hyperboloid of two sheets. 

2. ~^— Ti = 2cz, the hyperbolic paraboloid. 

-j;2 y2 

3. — 5 + r^ = 2c2, the elliptic paraboloid. 

128. Cylindrical Surfaces. If the circle 

x2 + 2/2*=25 (1) 

be moved parallel to itself so that all of its points describe lines 
parallel to the Z-axis, it will generate a right circular cylinder. 
The equation of this cylinder is sought. In any plane z = k, the 
relation between x and y for points in the curve of intersection of 
this plane and this cylinder is the same as that for points in the 
plane z = 0, viz., x"^ -{- y^ = 25. 

Now, this equation is satisfied by all points on the surface for 
all values of z. Hence it is the equation of the surface. 

The cylindrical surface just considered can be regarded as 
generated by a line moving parallel to the Z-axis and passing 
through points of the circle x^ -\- y^ = 25 in the plane z = 0. 

In general a cylindrical surface is a surface generated by a line 
moving parcdlel to itself. 



§128] SOLID GEOMETRY . 245 

It is clear that the equation 

Kx, y) = (2) 

represents the cylindrical surface generated by a line moving 
parallel to the Z-axis and passing through points of the curve 
f{x, y) = in the plane z = 0. The equation of a section of (2) 
made by any plane z = A; is j{x, y) = 0. 
Thus 

$ + $-^ (3) 

represents an elliptical cylinder whose elements are parallel to the 
Z-axis. 

3.2 _j_ 1^2 _ 2ax (4) 

represents a circular cylinder whose elements are parallel to the 
Z-axis. The center of the section in the plane z = is the 
point (a, 0). 

By the same reasoning 

7/ + z^ = a^ (5) 

represents a circular cylinder whose elements are parallel to the 
X-axis. 

z2 = 4x (G) 

represents a parabolic cylinder whose elements are parallel to the 
y-axis. 
The plane 

a; - 4y + 3 = (7) 

can be regarded as a cylindrical surface whose elements are 
parallel to the Z-axis and which pass through the line 

X , 3 

y-i + i 

in the plane z = 0. 

In general, an equation in which one of the letters x, y, z is 
absent, represents a cylindrical surface whose elements are 
parallel to the axis corresponding to the letter which does not 
appear in the equation. 



246 



CALCULUS 



I§129 



Exercises 

Describe the surfaces represented by the following equations: 
1. x^ + y^ = 16. 7. x« - 2/2 = 0. 



x^ V^ 
2- 4 +16 = 1- 

3. a;* - t/« = 1. 

4. 2» + y2 = 25. 

5. 22 - X* = 25. 
B. X + 2y = 10. 



8. a;?/ = 1. 

9. xz = 2. 

10. (x -3)(x + 2) = 0. 

11. y^ = 4x. 

12. 7/2 + 2" = 2ay. 

13. x2 + 2/2 = lOx. 



129. Partial Derivatives. Let z = f(x, y) be a function of two 
independent variables, x and y. When x takes on an increment 
Ax, while y remains fixed, z takes on an increment which we shall 

denote by AxZ. When y takes 
on an increment, Ay, while x 
remains fixed, z takes on an in- 
crement which we shall denote 
by A^2. 

For example, if a gas be en- 
closed in a cylinder with a mov- 
able piston, the volume v of the 
gas is a function of the tempera- 
ture T and of the pressure p 
which can be varied by varying 
the pressure on the piston. If 
the temperature alone be changed the volume will take on a 
certain increment A?^. If the pressure alone be changed the 
volume will take on the increment ApV. 

If 3 = /(x, y) be represented by a surface. Fig. 83, the increment 
of z obtained by giving x an increment, while y remains constant, 
is the increment in z measured to the curve cut out by a plane 
y = k, Q, constant. Thus A^z = HQ; similarly A^z = KR. 

AxZ 
The limit of the quotient -r— as Ax approaches zero is called the 

partial derivative of z with respect to x. It is denoted by the 

dz 
symbol -t^. Then 




dx 



dz 
dx 



lim AxZ 
^=0 Ax* 



§130] SOLID GEOMETRY 247 

It is evidently calculated from z = J{x, y) by the ordinary 
rules of differentiation, y being treated as a constant. Thus if 
z = x^y, 

dz 
Geometrically w— represents the slope of the tangent at the 

point (x, y, z) to the curve cut from the surface by the plane 
through this point parallel to the XZ-plane. 
Similarly 

dz_ _ lim A^ 

dy ^J/=o Ay 

and it is calculated by differentiating z = /(x, y), treating x as a 
constant. Geometrically it represents the slope of the tangent 
at the point {x, y, z) to the curve cut from the surface by the 
plane through this point, parallel to the FZ-plane. If z = xhj, 

■5— = x^. 
dy 

X 

Illustration 1. If 2 = sin -> 

y 



dz X d /x\ 1 X 

= cos - -^T" I - ) = - cos - 



dx y dx \yj y y 

and 

dz X d /x\ X X 

V- = cos - ^- ( - ) = ~ cos -• 

dy y dy \y} y^ y 

130. Partial Derivatives of Higher Order. If z is differentiated 
twice with respect to x, y being treated as a constant, the deriva- 
tive obtained is called the second partial derivative of z with re- 

dh 
spect to x. It is denoted by the symbol 3—^. Similarly the second 

partial derivative of z with respect to y is denoted by the symbol 

dy^' 

If z is differentiated first with respect to x, y being treated as a 
constant, and then with respect to y, x being treated as a constant, 

dh 
the result is denoted by the symbol „ „ • If the differentiation 

takes place in the reverse order the result is denoted by the symbol 



248 CALCULUS [§130 

- „ • The first is read "the second partial derivative of z with 
dxatj 

respect to x and y;" the second, "the second partial derivative 

of z with respect to y and x." In the case of functions usually 

occurring in Physics and Engineering, viz., functions which are 

continuous and which have continuous first and second partial 

d-z d'^z 

derivatives, „ ^ ■ = ^ „ • The order of differentiation is 
dyax oxay 

immaterial. 

Illustration 1. z = xhj. 

dz ^ d^z „ dh 

ai = 2^^^' d^^ = 2^' d^x = 2^- 

^ ^ 3.2 ^ = -^- = 2x 

^2/ ' dy^ ' dxdy 



In this case 



d^z _ d^z 

dydx dxdy 

X 



IlliLstration 2. z = 

dz 
dx " 


= sin • 

y 

1 X 

= - cos — 

y y 


dx^ " 


1 . X 

= 2 sin • 

y' y 


a?/ax 


1/ . x\ / x\ 1 X 
= - ( — sin I ( — „ ) — „ cos ~ 

y\ yJ \ yV 2/' 2/ 


= 


1 / . X x\ 
= , I a; sin w cos - ) • 

y'\ y ^ y) 


dz 

dy ' 


X X 

= i cos -• 

2/' y 


d^z 
dy'' 


2x X x' . X 

= 7 cos iSin -• 

2/^. 2/ 2/^ 2/ 


. dH 
dxdy 


a; / . a;\ /1\ 1 x 
= —i. I sm " I I - I i cos 

2/^ \ y) \yJ 2/' y 


= 


1 / . X x\ 
= —. I a; sin y cos - ) • 

2/' \ y y) 


Here again, we notice that 




d'z dH 
dydx ~~ dxdy 



§130] SOLID GEOMETRY 249 

Exercises 

d^z d^z d'^z d^z 

1. Find ^r^» ^r— ,' -> a ' and ^ a ' for each of the functions : 

(a) z = ^- (b) z = xy\ (c) z = xhj. 

(d) z = sin xy. (e) z = e* sin y. 

9^z 6^z 

2. Find , - - and - ^ for each of the following functions : 

dydx dxdy 

(a) z = x^y. (h) z = xsin'^y. (c) z = x cos y. 

Id) z = y log X. (e) z =- e" sin x. (J) z = y tan'i j;. 

It is seen that ^-^- = ^-^T in all of these cases. 
dydx dxdy 

In the above discussion z was considered to be a function of two 
independent variables only. The notion of partial derivatives 
can, however, be extended to functions of three or more variables. 

Illustration 3. li z = xhjt, 

ix = 2^^^' 

dy "" '' 
dt = ^ ^' 



and 



dy _^ 
dxdt " *'^^' 

d'z ^2x 
' dtdydx 



CHAPTER XV 

SUCCESSIVE INTEGRATION. CENTER OF GRAVITY. 
MOMENT OF INERTIA 

131. Introduction. In the preceding chapters there have been 
numerous examples of successive integration of functions of a 
single independent variable. Thus, to determine the law of 
motion of a falling body whose differential equation of motion is 

it is necessary to integrate twice. The result of the first integration 

ds 
is -^ = gt -{- Ci, and that of the second is s = ^gt^ + Cit + Cz. 

Exercises 

1. If -j—^ = 2x, find y as a function of x, given that -r- = 3 when 

X = 1, and y = 2 when x = 4; given that ?/ = 4 when x = 2, and 
that y = 7 when x = 4. 

d^y 

2. Find y if -7-^ = 7x. Assign suitable conditions to determine the 

constants of integration. 

3. Find y if ^ = 2x\ 

The operation of finding the result of Exercise 2 can be written 

J[ JiJtx dx\dx]dz = /[/[la;^ + Ci\dx\dx 

= ^ + Ci^- + C,x + C3. 
The first member can be written 

Ix dx dx dx. 



///' 



It is a triple integral and indicates that integration is to be per- 
formed three times in succession. An arbitrary constant of 

250 



§132] SUCCESSIVE INTEGRATION 251 

integration is introduced with each integration. If each integra- 
tion is performed between limits the constants of integration do not 
appear. Thus, 

7x dx dx dx = I I -^ dx dx 



is: 



42 dx dx 



= I 42 a; dx 



■r 



1 



= 1 84 dx 

|2 

= 84 X =84. 
|i 

132. Illustration of Double Integration. Let 

il = -' + y- <i) 

Integration with respect to y, treating x as a constant, gives 

^ = x^t/ + | + 0(x), 

where (t>(x) is an arbitrary function of x. This arbitrary function 
of X takes the place of an arbitrary constant of integration in the 
case of a single independent variable. A second integration, this 
time with respect to x, gives 

2 = ^ + ^' + J<A(^)rfx + ^(^), (2) 

where 4'{y) is an arbitrary function of y. 

The result contains an arbitrary function of x and an arbitrary 
function of y. Equation (2) represents a surface, but a very 
arbitrary one on account of the presence of the arbitrary functions 
I (f){x)dx and 4'(.y)- The process of finding (2) from (1) is indi- 
cated by a double integral sign. Thus, 



// 



(x2 + ?/2) dy dx, 



252 CALCULUS [§132 

which means 



I ( I (x2 + 2/2)di/) dx. 



Upon performing the integration indicated, first with respect to y, 
then with respect to x, we obtain 

I I (x2 + t/2) dydx = I (x^y +J + <f>ix)) dx 

Instead of an indefinite double integral such as the one just 
considered we may have a definite double integral. If the inte- 
gration with respect to y is performed before that with respect to 
X, the limits of integration with respect to y may be functions of 
X. Thus, 

I \\x^ + y') dydx=f (xH) + I) Y dx 

The last integration is readily j)erformed. It is to be noted that 
in evaluating a double integral, x is treated as a constant when the 
integration with respect to y is performed. 

If in a definite integral dx is written before dy, the integration 
with respect to x is to be performed first. ^ 



1. 1 1 xydydx. 


Exercises 

f*ir f*a{\ + C03 B) 

4. 1 1 rdrde. 

Jo Jo 


2. 1 1 xy dy dx. 

Jo Jo 


r- rva-^-x^. 

5. 1 dydx 

Jo Ja—x 


3. 1 1 xy-'dydx. 
Jo Jix 


6. 1 1 1 x^y^z^ dz dy dx. 
Jo Jo Jz 



1 Usage varies on this point. The student will have to observe in every case the 
convention adopted in the book he is reading. 



§133] 



SUCCESSIVE INTEGRATION 



253 



I dzdy 

Jo 

nVx n rvi 

xdydx. 10. I I 

n'' {x^ + y^) dy dx. 11- I I 

2 Jo Jx^ 



dx. 
10. 



12. 



dy dx. 



y dy dx. 



Vo^ — ^"^ — y^ dy dx. 



Hint. To perform the integration in Exercise 12, let v a^ — x^ = b 
and make use of the result of Illustration 3, §105. 

133. Area by Double Integration: Rectangular Coordinates. 
A plane area can be represented by a double integral. Thus, let 
it be required to find the area A between the curves y = /i(x), 
y = f^ix), and the lines x = a and x = b. The area of the strip 
IJKH, Fig. 84, is approximately 



lim 

Ay 



«2 «t /.V, 

j2)At/Aa; = Aa;]^m^2)Ay = Ax I dy, 



where yi and y^ are the ordinates of the two curves y = /i(x) and 
y = fiix), respectively. And 
the area sought is approxi- 
mately 



x = a Jvi 




Fig. 84. 



The smaller Ax is taken, the "o 
closer the approximation. 
The limit of this sum as Ax 

approaches zero is the area sought. Since y\ and 2/2 are func- 
tions of X, 1 dy is, a, function of x and consequently 






lim 



K)X Ax I ^2/= I \ dy 



dx = A. 
It is to be noted that in setting up this integral the summation 



254 



CALCULUS 



[§133 



with respect to y was performed first, giving the area of a vertical 
strip for a particular value of x. Consequently, the integration 
with respect to y is to be performed first, x being treated as a 
constant. On performing the integration with respect to y we 
obtain 

(2/2 - yi)dx = I [/zCx) - Si{x)\dx, 

a Ja 

a single integral which might have been set up at once by consider- 
ing the area as the sum of vertical strips of length 1/2 — yi, and 
of width Ax. It is, however, desirable to be able to set up a 
double integral over an area. 

In choosing the limits for a double integral, the student should 
proceed systematically. The process of setting up the above 

integral with its limits is the follow- 
ing: The "element" is the rectan- 
gular element of area dy dx. The 
"summation" (integration) of this 
element, for a particular value of x, 
between the limits for y of WI and 
WH, the ordinates of the curves y = 
/i(x) and y = /2(.t), gives the area of 
the typical strip IJKH. The "sum- 
mation" (integration) of the strips of 
which this is a typical one, between 
the extreme values oi x, x = a and x = b, gives the area sought. 
Thus 

A = I I dy dx. 




nv 



The procedure may be briefly summarized in the following concise 
directions. Write first the element dy dx, then the integral sign, 
then the limits /i(x), /2(x), then another integral sign with the 
limits a and b. 

Illustration. Find by double integration the area between the 
parabolas y^ = x and y = x^. The integral is set up as follows: 
Write the element dy dx, then an integral sign with the limits 
x^ and \/x. This represents the area of the typical strip, IJKH, 
Fig. 85, for a fixed x. All of the strips of which this is a typical 



§134] SUCCESSIVE INTEGRATION 255 

one are to be summed from a; = to x = 1, the abscissas of the 
points of intersection of the curves. Then write the second 
integral sign preceding the first with the limits and 1. Thus 



Jo Jx' 



dy dx = \. 



Exercises 

1. Find by double integration the area between the curves y = x 
and y^ = x'. 

2. Find the area of Exercise 1 by integrating first with respect to x 
and then with respect to y. 

3. Find by double integration the area between r/* = a{a — x) and 
X* + 2/^ = a^. 

4. Find the area between y^ = ax and y^ = 2ax — x^. 

5. Find the area of Exercises 3 and 4 by integrating first with respect 
to X. 

6. Find the area bounded by y^ = 4x, x + y = 3, and the X-axis. 

7. Find the area of a rectangle by double integration. 

8. Find the smaller area between x^ + y^ = 1 and y = x + 5- 
134. Geometrical Meaning of the Definite Double Integral 

Consider the definite double integral 

fix, y) dy dx. (1) 

, = /i(x) 



J'»6 /»2/2=/20 



In accordance with the definition of a definite single integral, 
§66, (1) can be written 

f'[ll^oXf(^'y) ^y]d^' (2) 



Here x is considered constant under the summation sign, and 
f{x, y) is, for such a fixed x, a function of y alone. 

V, 

lim 



^2 /»2/2 



is a function of x, since x occurs as an argument of / and also in 
the limits of integration. Hence we can write (2) in the form 



256 



CALCULUS 



[§134 



lim 



O L «. J 



-E%'Zll%Xn-.y)^vAx, 



(3) 



where Ax under the second summation sign is regarded as a con- 
stant multiplier. 

Z 




Fig. 86. 

In Fig. 86, let EFGL represent the surface z = f{x, y) ; QABS, 
in the ZF-plane, the curve y = fi{x) ; DHKC, the curve y = fi{x) ; 
AD the line x = a; BC the line x = b; and A'B'C'D' the portion 
of the surface cut from z = f(x, y) by the cylinders y = fi{x), 
y = f^ix), and the planes x = a and x = b. 



§134] SUCCESSIVE INTEGRATION 257 

Divide A BCD into small rectangles, as shown in the figure, by 
lines parallel to the X- and F-axes, at intervals of At/ and Ax, 
respectively. Through these lines pass planes parallel to the 
XZ- and FZ-planes. These planes divide the solid bounded by the 
planes and surfaces of Fig. 86 into vertical columns of rectangular 
cross section Ay Ax. The column erected on MNPR as a base is a 
typical one. f{x, y)Ay Ax represents approximately the volume of 
the column whose base is MNPR and whose top is M'N'P'R', since 
the area of its base is Ay Ax and its altitude is MM' = f{x, y). 
Then the sum of the columns at a fixed distance from the 
yZ-plane, 

2)/(x, y) Ay Ax, 

is approximately the volume of the slab between the planes 
IHH'I' and JKK'J', i.e., between the planes x = x and 
X = X + A X. And 

h 



r «2 -1 

a L w. -I 



the sum of the volumes of all the slabs, is approximately the vol- 
ume of the solid ABCDA'B'C'D'. If Ay and Ax are each taken 
smaller and smaller this sum will eventually represent a very close 
approximation to the volume in question, and the limit of this 
sum as Ay and Ax approach zero is the volume. Hence the inte- 
gral 

f(x, y) dy dx, 



which we have seen is equal to 

X = b r- V 



a; = a L v. J 



lim 

x = a 



represents the volume bounded by the plane z = 0, the surface 
^ = /(-^j y), the planes x = a and x = h, and the cylinders 
y = /i(.t) and y = Jiix). 

Illustration. Find the volume contained in the first octant of 

17 



258 



CALCULUS 



[§135 



the sphere x^ + y^ + z"^ =- a^. See Fig. 87. The equation of the 
surface is 

z = \/a2 — x^ — y^. 

yi = fiix) = 

2/2 = f2{x) = Va^ - x\ 

the trace of the sphere on the XF-plane. The volume of the col- 
umn on MNPR as a base is 

Va^ — x^ — 2/2 A?/ Ax, 

or, as we shall say in the future, 

\^a^ — x^ — y^ dy dx. 

The summation of these columns 
for a fixed x gives 




Fig. 87. 



r 



■y/a'^ — x^ — y"^ dy dx, 



!/■ 



the volume, expressed as a function of x, of the slab between the 
planes x = x and x = x4-Axorx = x + dx. The summation 
of all these slabs from x = to x = a gives 

*y/a^ — xi 

^/a^ — x^ — y^ dy dx, 
to 

the volume of one octant of the sphere. This integral was evalu- 
ated in Exercise 12, §132. 

Exercises 

1. Find the volume of the segment of the paraboloid y^ + 2z^ = 4x, 
cut off by the plane x = 5. 

2. Find the volume bounded by the cylinders y = x^ and y- = x, 
and the planes z = and z = 1. 

3. Find the volume common to the cylinders x^ + y^ = a^ and 
y^ + z^ = a2. 

4. Find the volume between the cylindrical surface y^ — x', the 
plane y = x, and the planes z = and z = 1. 

135. Area: Polar Coordinates. Let it be required to find, by 
double integration, the area between the radii vectores 6 = a, 



§135] 



SUCCESSIVE INTEGRATION 



259 



and d = ^, and the curve p = f{d). Divide the area as shown in 
Fig. 88, the radii making an angle of Ad with each other and the 
radii of the concentric circles differing by Ap. The area of MPQR 
is equal to 

Kp + Ap)2 A0 - § p2 A^ = p Ap A0 + ^ Ap2 A0 

As Ap approaches zero, 

limpApA0 + |Ap2A5 



Hence 

P=Ke) 



^"=0 p Ap A^ 



= 1. 



p = p = Jp = 





Fig. 88. 



Fig. 89. 



This sum represents the area of the sector OHL. The total area 
sought is the limit of the sum of these sectors as A9 approaches 
zero, i.e. 

^ = Ai"oX^^ p^p= \ p^p^- 



This integral is to be set up as follows: The element of area is 
the approximately rectangular area MPQR whose area is approxi- 
mately {MR){MP) = pdpdd. This element is to be summed 
from p = to p = f{d) . This gives approximately the area of the 
typical sector OHK. These sectors are to be summed from 
e = atod = fi. 



260 CALCULUS [§136 

More briefly: Write down the element pdpdd, then an 
integral sign. Its limits are the extreme values of p for a given 6. 
Then write before this integral another integral sign. Its limits 
are to be chosen so as to sum up all the sectors such as OHK. 

Illustration 1. Find the area 
of the circle p = 10 cos 6, Fig. 
89. The area bounded by the 
semicircle above the initial line 
will be found and multiplied by 
two. 



.20 cos ff 




Jo Jo 



lOcosO 

A =2\ I pdpde. 



p dd dp. 

Illiistration 2. Find by double integration the area between 
p = 10 cos e and p = 20 cos d. See Fig. 90. 



•20 COS e 
A =2\ I pdpdd 

'l0CO3 tf 

Show that 



t/O t/10 

mo /•«=o«-'^ /•20 /-cos-^ 

A = 2 I I ^ pdddp+2 \ I pdddp. 

Jo JC08-1-^ JlO Jo 



10 

Which method is the simpler in this case? 

Exercises 

Find by double integration: 

1. The area of the circle p = 5 sin 0. 

2. The area of the cardioid p = a(l — cos 9). 

3. The area of the lemniscate p" = a" cos 20. 

4. The area outside p = a (1 + cos 6) and inside p = 3a cos d. 
136. Volume of a Solid: Triple Integration. We shall now 

find the volume of the solid of Fig. 86 by triple integration. Sup- 



§136] SUCCESSIVE INTEGRATION 261 

pose the solid further subdivided by planes parallel to the XY- 
plane and at a distance Az apart, into rectangular parallelopipeds 
of Volume Az Ay Ax. Then the volume of the column on the base 
MNPR is approximately 

a = f(.x, y) 

^ Az Ay Ax. 

Z = 

Then 

Vi z => f(.x,y) 

'^ 2^ Az Ay Ax 

y\ « = 

is approximately the volume of the slab between the planes 
IHI'H' and JKJ'K', i.e., between the planes x = x and 
X = X ■\- Ax. 

X = h 2/2 2 = f{x, y) 

X 2) 2) Az Ay Ax 

X = a y^ « = 

is approximately the sum of the volumes of all of the slabs. 
If Az, Ay, and Ax are each taken smaller and smaller, this sum will 
represent a very close approximation to the volume sought, and the 
limit of this sum as Ao:, Ay and Az approach zero is exactly this 
volume. Hence the integral. 



a Jy^ Jo 



dz dy dx, 



represents the volume bounded by the plane z = 0, the surface 
z —j{x,y), the planes x = a and x = h, and the cylinders 
y = j\{x) and y = Si{x). 

Illustration 1. Find by triple integration the volume of the 
ellipsoid 

a" ^ 62 -r g2 
See Fig. 91. 

F = 8 I I dzdydx 

Jo Jo Jo 



262 



CALCULUS 



[§136 



The student will perform the integration. 

Illustration 2. Find by triple integration the volume of the 
solid bounded by the cylinder x^ -\- y^ = 2ax, the plane z = 0, 
and the paraboloid of revolution x^ -{- y^ = ^az. Write the ele- 
ment of volume, dz dy dx. The integration with respect to z 

cc^ "4* 1/^ 
-J gives the volume of the typical 



between the limits and 



vertical column of base dy dx, extending from the point (x, y) in 

/j;2 _|_ y2 

the plane z = to the surface of the paraboloid, z = — j ' 

Next, X being kept fixed, these columns are summed into a 
typical slab by integrating with respect to y from the Z-axis, 




y = 0, to y = ^/2ax — x^, the trace of the cylinder in the XY- 
plane. Finally the integration with respect to x from x = to 
X = 2a gives one-half of the total volume sought, viz., that lying 
in the first octant. 

xi + y^ 

F = 2 I I I dzdy dx. 



Jo Jo Jo 

The student will perform the integration. 



Exercises 

1. Find the volume common to the cylinders x" + y^ = r^ and 

x2 -f- 22 = r^ 



§137] SUCCESSIVE INTEGRATION 263 

2. Find the volume of one of the wedges cut from the cylinder 
^2 ^ y2 — J.2 \yy ^hc plancs z = and z = mx. 

3. Find the volume in the first octant bounded by the coordinate 

X y z 
planes and the plane — h r H — = 1. 
^ ^ a c 

4. Set up the integral representing the volume bounded by the 
surface x^ -\- y^ + z^ = o*. 

6. Find the volume between y^ + z^ = 4ax and x — z = a. 

6. Find the volume between the planes y = 0, z = and the sur- 
faces z = x^ + Ay^, y = 1 — x'^. 

7. Find the volume between y^ + 2z^ = 4x and z = x. 

137. Center of Mass, Centroid. Let there be a system of 
masses mi, m2, ms, . . . , rUn situated at the points {xi, yi, Zi), 
{xi, 2/2, 22), (^3, 2/3, Zz), . . . , {Xn, yn, Zn) , Tespectlvely. The mean 
distance with respect to mass, of the system from the FZ-pIane is 

The mean distances, with respect to mass, of the system from the 
ZX- and XF-planes are, respectively, 

2 m.i/< 



and 




(2) 



(3) 



The point (x, y, z) is called the centroid, or the center of mass, of 
the system of masses mi, m2, • • • , m„. 

m,- Xi is called the moment, and Xi the moment arm, of the mass 
mi with respect to the FZ-plane.^ Then x is the mean moment 
arm with respect to the FZ-plane of the masses mi, m2, • • •, m„. 
For, equation (1) shows that if all the masses were placed at the 
distance, x, from the FZ-plane, the moment with respect to this 
plane would be the same as the sum of the moments of the masses. 
Hence we can say that the centroid of a system of masses is a point 
such that if all the masses were concentrated at this point, the 
moment with respect to each coordinate plane would be equal to 

1 The term moment of a mass with respect to a plane has evidently a diflerent 
significance from the term moment as applied to a force. 



264 CALCULUS [§139 

the sum of the moments with respect to the corresponding planes 
of the masses in their given positions. 

138. Centroid Independent of the Position of the Coordinate 
Planes. It will now be shown that the distance of the centroid, 
(x, y, z), from any plane is the mean of the distances, with respect 
to mass, of the masses mi, m2, • • •, m„, from that plane. And 
thus it will be shown that (x, y, z), the centroid, is a point whose 
position with reference to the masses is independent of the choice 
of the coordinate planes. 

Let ax -\- hy -\- cz -\- d = hQ the equation of a given plane 
(see §120). The distance, p, of the point (x, y, z) from this plane 
is 

- ax + hy + cz ■\- d 

P = 22 ' ^^^ 

where 

R = Va^ + 62 + c\ 

(See §124.) On substituting the values of x, y, and 2 from (1), 
(2), and (3) §137, and reducing the absolute term, d, to the common 
denominator, we have 

aSwiX,- + blfTTiiyi + cZniiZi + dXnii 



or 



sm.r 



R'Emi 
axi + hyi + czi + d' 



R J . (2) 

ox,- + hyi •\-czi + d • +1 j- . f .1 • ^ 

But 5 = pi is the distance of the point 

K 

(x,-, yi, Zi) from the given plane. Hence (2) can be written 

in the form 

_ S mipi 

This proves the statement at the beginning of this section. In 
other words, if all the masses of the system were concentrated at 
the centroid, the moment with respect to any plane would be 
equal to the moment of the system of masses with respect to this 
plane. 

139. Center of Gravity. Let the system of masses considered 
above be acted upon by gravity. It will be shown that the line 



§140] SUCCESSIVE INTEGRATION 265 

of action of the resultant force passes through the center of mass, 
or the centroid. 

Since the position of the centroid is independent of the choice of 
axes, choose the positive direction of the axis of z vertically upward 
and the axes of x and y horizontal. The force acting on mi is 
mi^, that on m^ is m-ig, etc. The resultant force is equal to 2 niig 
and is directed vertically downward. Its line of action meets the 
XF-plane in a point (a, j3, 0) such that its moment, aSm.gr, about 
the F-axis is equal to 2m,gfXi, the sum of the moments of the 
forces acting on the individual masses; and such that the moment, 
jSSmigr, about the X-axis, is equal to the sum of the moments, 
Sw.grr/,, of the forces about this axis. 



and 
Whence 

and 

Consequently 



Hence the resultant passes through the centroid of the system of 
masses. 

If the masses all lie in one plane, say the XF-plane, z is zero and 
the centroid is fixed by the two coordinates z and y. The product 
rtiiXi is called the moment of the mass rrii with respect to the 
F-axis. In this case x is the mean moment arm with respect to 
the F-axis. 

If the masses all lie upon a line, say the X-axis, the centroid is 
fixed by a single coordinate, x. 

140. Centroid of a Continuous Mass. If instead of discrete 
masses we have a continuous mass, the coordinates of the center 
of mass, or the centroid, are clearly. 



a2,m 


iQ = 'Lrrii 


g^i, 


/SSw 


xQ = 'Lrui 


gvi- 




^rUiXi 




Oi = 


^rrii 




^ = 


^rmyi 




2)^< 




a = 


X and |8 


= y- 



266 CALCULUS [§140 



Ani = 



X = 



y = 



i%^Am Jdm 



z = 



IL^.oX^ /dm 
^jj Vz Am J z dm 
^ X -^m fdm 



lim 

Am 



Am=( 

The integration is to be extended throughout the entire mass, and 
the integrals considered may be single, double, or triple, depending 
on the form of the mass. 

Illustration 1. Find the center of gravity of a bar, Fig. 92, 
of length L, whose linear density, p, may vary. Let the axis of x 
coincide with the bar, the origin being taken at one end. The 



OF 



^1 



Fig. 92. 

mass of an "element" of the bar of length dx is p dx, p being a 
function of x, the distance of the element from the origin. The 
moment of this element of mass, dm = p dx, about an axis 
through the origin perpendicular to the bar is 

X dm = xp dx. 

X, the abscissa of the centroid, the only coordinate necessary 
to fix the centroid in this case, is given by 

px dx 

mJ 

X = 



f. 



p dx 



The numerator represents the total moment, and the denominator 
the total mass. If the bar is of uniform density, p can be taken 



il40] 



SUCCESSIVE INTEGRATION 



267 



out from under the integral sign. Then 

xdx — 
2 

X = 



r 



^L. 



dx 



x\ 



If the linear density is proportional to the distance from one end, 
then p = kx and we have 

k \ x^ dx 1 



f 



k 



f. 



X dx 



Illustration 2. Let it be required to find the center of gravity 
of a plate of uniform thickness and of mass p per unit volume or of 
mass p per unit surface. Take a plate of the shape of Fig. 84. 
The mass of the element MNPR is p dy dx. The moment of this 
element about the Y- axis is xp dy dx, and its moment about the 
X-axis is jj p dy dx. Then 



r r 



p X dy dx 



fdm r p p^y^^ 



p y dy dx 



If p is constant, 



and 



i ^^« r r^pdydx 

Ja Jvi 

I xdy dx 

I { dy dx 

Ja Jy^ 

J'*b nvi 
I ydydx 
a Jy^ 

y^ rb /»!/, ' 

II dy dx 

Ja Jv^ 



268 CALCULUS [§140 

The numerator of each of these expressions is the integral of the 
product of an element of area by its distance, x or y, from the Y- 
axis or X-axis, respectively. The denominator is the area. The 
mass does not enter into either of these formulas. We are thus 
led to speak of the centroid of an area, of a line or of a solid, with- 
out reference to its mass. This notion of the centroid of a geo- 
metrical figure, a line, an area, or a solid, without reference to its 
material composition is an important one. For, in many prob- 
lems in mechanics one is interested in the centroid of a geomet- 
rical configuration as such. Thus in the study of the deflection 
of beams it is necessary to know the position of the centroid of the 
cross section of the beam. 

Illustration 3. Find the centroid of the solid represented in 
Fig. 86. The element of mass, dm, is equal to p dz dy dx, and its 
moment with respect to the yZ-plane is xp dz dy dx. Then 

Vix, y) 

px dz dy dx 



I X dm 
I dm 



Ja Jy^ Jo 



Similarly : 



Ja Jy^ Jo 
J Jo 



'fix, y) 

p dz dy dx 



'/(x, y) 

py dz dy dx 



and 



_ \y dm 

J^^ I Pdzdydx 

Ja Jvi Jo 



j zdm 
1 dm 



rb ny, n 

Ja Jvi Jo 



'Kx, v) 

pz dz dy dx 



rb ry, r 

Ja Jy^ Jo 



iix, y) 

p dz dy dx 



If the density is constant, p can be canceled from numerator 
and denominator. 

If the solid has an axis or a plane of symmetry the centroid lies 
in this axis or in this plane. 

Illustration 4. Find the centroid of the area in the first quad- 
rant bounded by the circle x^ -\- y^ = a^. 



§140] SUCCESSIVE INTEGRATION 269 

If we use double integration we have, in accordance with Illus- 
tration 2, 

X dy dx 



n 

— Jo Jo 



4 
and 

ydydx 

y = — "—^ 

4 
Radicals could be avoided in the evaluation of the numerator 
of the expression for x if the integration were performed first with 
respect to x and then with respect to y. Thus 

"Va2 - j/2 

X dx dy 

'0 «/o 
X = ^ • 

1" 

The student will evaluate each expression given for x. 

From the symmetry of the figure, x = y, and it is not necessary 
to evaluate the integral for y. 

In finding the centroid in this case, and indeed in many cases, 
it is easier to use single integration than double integration. 
Thus if we choose as the element of area, the strip y dx parallel 
to the y-axis, the moment of this strip about the F-axis is xy dx, 
and 

X dm I xy dx I x\/a' — x^ dx 
Jo Jo 



h f 



ydx -^ 



Illustration 5. Find the centroid of the solid in the first octant 
bounded by the sphere x^ -\- y^ -\- z^ = a^. 
The method of Illustration 3 gives 

I X dz dy dx 

Jo Jo Jo ' 

X = -, 



270 CALCULUS (§140 

From considerations of symmetry, y = z = x. 

Here again it is simpler to use single integration. Choose as 
element a slab of thickness dx parallel to the FZ-plane. The 
base of such a slab is a quadrant of a circle of radius \/a^ — x^, 
where x is the distance of the slab from the FZ-plane. The volume 
of this elementary slab is 



7r(a^ — x^) , 
z ax. 



Hence 

TT 



X = 



I X {a^ 



. , .^ . V. x^) dx 
4. f ^ 



Exercises 

Find the coordinates of the centroid of: 

1. The area between y = x^ and y^ = x, 

2. The areas of Exercises 1, 3, and 6, §133. 

3. A triangular plate. 

Hint. Draw lines parallel to the base, BC, Fig. 93, at intervals 
dx along the median AM. The mass of each strip is proportional 

to AL = X and can be regarded as 
concentrated at its centroid on the 
line AM. Hence we can think of 
the triangular plate as replaced by 
the bar AM whose density is propor- 
tional to the distance from the end 
A In accordance with Illustration 
1, its centroid is at a point / two- 
FiG. 93. thirds of the way from A to M. 

The centroid of a triangle can also 
be located without any calculation whatever. From Fig. 93 it follows 
that the centroid lies on the median AM. The same argument shows 
that it lies on the medians BMi and CM2. Hence it lies at the 
point of intersection of the medians, i.e., at a point two-thirds of the 
way from a vertex to the middle of the opposite side. 

4. The area of a semicircular plate of radius r. (Single integra- 
tion will be sufficient.) 




jl40] 



SUCCESSIVE INTEGRATION 



271 



5.1 Let OMKB, Fig. 94, be a quadrant of a circle of radius r. Let 
OMDB be a square. Denote by Ci, C2, and C3 the centers of gravity 
of the square, the quadrant of the circle, and the area MDBKM, 
respectively; and by Ai, A2, and A3 the corresponding areas. Then 

^2X2 + AzX3 = AiXi 

A2X2 



AiXi 
X3 = 



= 0.223 r. 



DC3 = 0.315 r. 
6. A circular arc of radius r and central angle 2a. See Fig. 95. 
















Ai 


\ 




/ 


/ 






\ 


\"~ 


/C2 






*- \ 








V 










/ 


-^^ 








D 








t 


i 




Fig. 94. 



Fig. 95. 



Hint. The centroid lies on the radius which bisects the central 
angle since this line is an axis of symmetry. Choose this radius as the 
axis of X and the center of the circle as the origin. Then y = 0, and 



- Xr! X 

^ ~ Ira. ~ 



r cos Or do r'- 

— a 

' Ira. 



/:.• 



cos do 



2rc 



This problem can also be solved by using rectangular coordinates. 
Thus 



— t/r C( 



xdx 

Wj.2 — 



2rc 



2r^ sin a r sin a 
2ra 



7. The portion of the arc of the circle x^ + j/^ = r^ which lies in the 

1 Exercises 5, 9, 20, 21, and 22 taken from Technical Mechanics by Maurer. 



272 



CALCULUS 



[§140 



first quadrant. Use the result of Exercise 6. Also find the result 
directly. 

8. The parabolic segment of altitude a and base h. See Fig. 96. 
Hint. Show that the equation of the parabola is ^ay"^ = b^x. 

Ans. X = fa. 

9. A conical or pyramidal solid of altitude a and base A. 

Hint. Let OMNO, Fig. 97, represent the projection of the solid 
on the XF-plane. Divide the solid by planes parallel to the base 
into laminas or plates of thickness dx. Then the area of the lamina 





' 










Y 






N 






^.---^ 




~" 


" 




< 


a >^> 




o 


f 


a 






' K 






^ 


/ 


/^ 






\ 










^^...^M 




\ 


^--^^ 




\ 


, 


O 


^ 


•"■""'''^ ! ! X 








dx 






Fig. 96. 












Fig. 


97 


, 





Ax"^ 



whose abscissa is x is — ^: and its volume is 
solid is -0-. Hence : 



Ax^dx 



The volume of the 



J'*'' /Ax^dx\ 
X = 1 



Aa 
T 



3a 
4" 



Further the centroid of every lamina lies on the line joining the apex 
with the centroid of the base. Consequently the centroid of the 
solid lies on that line. 

10. The hemisphere generated by revolving one quadrant of 
x» -(- j/2 = r^ about the A'-axis. Evidently 7/ = 2 = and 



■I 



xy' dx 



11. The surface of the hemisphere of Exercise 10. 

12. The segment of a paraboloid of revolution of altitude h. 



§140] 



SUCCESSIVE INTEGRATION 



273 



13. The semi-ellipsoid of revolution generated by revolving one 
quadrant of -^ + r^ = 1 about the X-axis. 

14. The surface of the paraboloid of Exercise 12. 

15. The surface of a right circular cone. Any conical or pyramidal 
surface. 

16. The area of the cycloid a; = a{d — sin e),y = a(l — cos 6). 

17. The arc of the cycloid of Exercise 16. 

18. The area in the first quadrant under x^ + y^ = a^. 

19. The arc of the curve of Exercise 18 in the first quadrant. 

20. The segments of the ellipse indicated in Fig. 98. It will be 
found that the centroid of the segment XAAX coincides with that of 




Fig. 98. 



the segment XaaX of the circumscribed circle, and that the centroid 
YBBY coincides with, that of the segment YhhY of the inscribed 
circle. 

21. Ci and Cj are the centers of gravity of the two portions of Fig. 
99. Show that their distances from the sides of the enclosing rec- 
tangle, a X b, are those marked in the figure. The curve OC is a 
parabola. See Exercise 5. 

22. Find the centroid of the portion of a right circular cylinder 
shown in Fig. 100. C is the centroid. Its distance from the axis of 

the cylinder shown is — jr — -, and from the base is « H pj, — 

When the oblique top cuts the base in a diameter (lower part of Fig. 

100) the distance of the centroid from the axis is -r^ and from the base 



16 



3ira 
■32' 



18 



274 



CALCULUS 



[§141 



23. Find the centroid of the volume lying in the first octant and 
included between the cylinders x- + y^ = a^, x^ + z* = a^ 



4«H 




\ 




C 

• 




1 

1 

" "- 1 


■ > 

a 

__J_ 


^^ ,C 



Fig. 99. 



Fig. 100. 



141. Theorems of Pappus. Theorem I. The area oj the surface 
generated by revolving an arc of a plane curve about an axis in its 
plane and not intersecting it is equal to the length of the arc multi- 
plied by the length of the path described by its centroid. 

Theorem 11. The volume of the solid generated by revolving a. 
plane surface about an axis lying in its plane and not intersecting 
its boundary is equal to the area of the surface multiplied by the 
length of the path described by its centroid. 

Proof of I. Let ABC, Fig. 101, be an arc of length L lying in the 
XF-plane. Then y, the ordinate of its centroid, is given by the 
equation : 



H 



M 



Whence 






Fig. 101. 



/ 



yds = y L. 



(1) 



(2) 



The surface generated by revolving the arc ABC about the 
X-axis is given by 



§141] SUCCESSIVE INTEGRATION 275 



/ 



S = 2Tr \ yds. (3) 

It follows then from (2) and (3) that 

S = 2TryL. (4) 

But 2iry is the length of the circular path described by the cen- 
troid of the arc ABC. Hence the theorem is proved. 

Proof of II. Let ABC, Fig. 102, be a plane surface of area A. 
Then y, the ordinate of its centroid, y 
is given by 



. / 



ydA 

-y = ^-j- (5) 

Whence ^ 



/ 




ydA=Ay. (6) Fig. 102. 



Now the volume of the solid generated by the revolution of the 
area ABC about the X-axis is 

V = 2ir I j ydydx = 2t I y dA. (7) 

It follows from (6) and (7) that 

V = 2TrAy. (8) 

Hence the theorem is proved. • 



Exercises 

1. Find the surface of the anchor ring generated by revolving the 
circle x'^ + {y — b)^ = a^, a < b, about the X-axis. 

2. Find the volume of the anchor ring of Exercise 1. 

3. Find, by using one of the theorems of Pappus, the centroid of a 
quadrant of a circular arc, radius a. 

Hint. The rotation of the arc about the X-axis, which coincides 
with a radius drawn to one extremity, generates the surface S = 2ira'^. 
Then, by (4), 

S = 27ra2 = 2T2/L = 2-Ky'^' 



276 



CALCULUS 



[§142 



Hence 



y 



2a 



4. Find, by using one of the theorems of Pappus, the centroid of a 
quadrant of a circular area. 

142. Centroid: Polar Coordinates. The formulas are readily 
obtained for finding the coordinates of the centroid of an area 
bounded by a curve whose equation is given in polar coordinates. 
The area of the element MPQR, Fig. 88, is p dp dd, and its moment 
about the F-axis is pdpdd p cos 6 = p"^ cos d dp dd. Hence 



Similarly, 



fCp"^ cose dp do 
ff pdpdd 

f fp^ sine dp dd 
ff pdpdd 



X = 



y = 



If it is advantageous, the integration with respect to d can be 
performed first. 

Exercises 

Find the coordinates of the center of gravity of: 
1. The area of p = a(l + cos 6). The area of the upper half of 
the same cardioid. 

2. The area of one loop of p = 
a cos 2d. 

3. A circular sector of central angle 
2 a. 

4. One quadrant of a circle. A 
semicircle. (Obtain directly and also 
use the result of Exercise 3.) 

5. The area of a portion of a cir- 
cular ring, Fig. 103, of radii R and 
r, and of central angle 2a. Denote 
by Cr the centroid of the sector of 

radius R, and by Cr that of the sector of radius r, and by C that 
of the given portion of the ring. Let the abscissas of these points be 
xr, Xt, and x, respectively, and let the corresponding areas be denoted 
by Ar, Ar, and A. 




Fig. 103. 



§143] SUCCESSIVE INTEGRATION 277 

Then 



Hence 



Artr ■\- Ax = ArXr. 

- _ "^g^fl - ^rXr ^ 2 R^-r^ sin a 
^~ A 3 R^-r^ a ' 



Obtain this directly by integration. 

6. A segment of a circle of radius r cut off by a chord of length c. 
Use the method of Exercise 5. The distance of the centroid from the 
center is 

(.3 ^ 2r^ sin^ a 

12A ~ M ' 
where A = area of the segment = r^ (2 a — sin 2 a:). 

143. Moment of Inertia. Consider a system of masses, mi, 
rriz ■ ■ . , rrin, moving with linear accelerations, ji, j^, • • ■ , jn, 
respectively. The forces acting on these masses are then m\j\, 
m^ii, . . ., rrinjn, respectively; and the sum of the moments of 
these forces about an axis is equal to l^niijiri where ri, rz, . . . 
r„, respectively, are the moment arms of these forces with respect 
to this axis. If now the masses are rigidly connected and rotate 
about an axis, they have a common angular acceleration. Let the 
common angular acceleration be denoted by a. Then ^i = ari, 
ji = ar2, . . ., jn = oLTn, whcro ri, r2, . . ., r„ are the dis- 
tances of the masses mi, m^, . . ., rUn from the axis of rota- 
tion. The sum of the moments, T^mijiTi, becomes a'Zmiri^. 
This is the moment necessary to produce the angular acceleration 
a. To produce unit angular acceleration a moment equal to 
Sm.ri^ is necessary. This moment, ^miri"^, is called the moment 
of inertia, and is denoted by the symbol /. Thus 

I = SmiV. (1) 

The moment of inertia of the system would be unchanged if the 
n masses of the system were situated at a distance k from the axis 
of rotation such that 

'Zniik'^ = 'ZmiTi^, 
or 

^ ~ 2mi 



278 



CALCULUS 



[§143 



k is called the radius of gyration. Its square is the mean of the 
squares of the distances ri, ri, . . . , r„ with respect to the mass. 
The moment of inertia of a system with respect to an axis of 
rotation plays the same role in the discussion of a motion of rota- 
tion as the total mass in the discussion of a motion of translation. 
In the former case the moment necessary to produce an angular 
acceleration a is a^niiri^. In the latter case the force necessary 
to produce a linear acceleration j is jSm,-. 

The kinetic energy of a rotating system can be expressed in 
terms of its moment of inertia and its angular velocity. If a 
particle of mass m is rotating with angular velocity co about an 
axis and at a distance r from it, its kinetic energy is equal to 

one-half the product of its mass 
by the square of its linear veloc- 
ity, i.e., to jWwV^. And if 
there is a system of particles of 
masses mi, m2, • • • , w„, at dis- 
tances ri, r2, . . ., r„, respec- 
tively, from the axis, all rotating 
with the angular velocity w, the 
kinetic energy of the system is 
equal to ^SmiCoVi^ = ^oo^Ziriiri^ 

= Wl- 
If a rectangular plate of uniform thickness ^ and composed of 
material of uniform density, p, rotate about an axis through one 
corner and perpendicular to its plane, its moment of inertia can be 
found by a process of double integration. Let the sides of the rec- 
tangle be a and h and take the origin at one corner. Fig. 104. 
The moment of inertia of the rectangle MNPQ is approximately 
the product of its mass, p^ Ay Ax, and the square of the approxi- 
mate distance, s/x^ -\- y"^, of its mass particles from the origin. 
That is, the moment of inertia of MNPQ is approximately 

p^(x2 -I- y^)Ay Ax. 
That of the strip EHIJ is approximately 

y = b ^^ 

P^ AySo DC-^' + y') AyAx=p^Ax i (x' + y') dy. 

And the moment of inertia of the entire plate is obtained by 



y 








J 


I 
















































Q 




P 
























W 
























M 




N 
































































X 


o 








B 


7 J 


1 















Fig. 104. 



§143] SUCCESSIVE INTEGRATION 279 

taking the limit of the sum of the moments of inertia of these 
strips as Ax approaches zero, viz., 



«/o Jo 






I = P^£to^^x\ {x^ + y')dy 



b 

(x2 + 2/2) dy dz = Wia-" + 6^), 



where M — p^ab, the mass of the plate. 

We have obtained the moment of inertia of the plate by inte- 
grating over its area the product of the mass of the element, 
p^ dy dx, by the square of its distance, ^Jx^ + y"^, from the axis of 
rotation. 

If instead of a rectangular plate we consider a plate of any shape, 
say that of Fig. 84, its moment of inertia is given by 

(x^ + y^) dy dx. (2) 

If the density, p, and the thickness, ^, are variable the foregoing 
argument shows that they must be written under the integral sign. 
For, the element of integration is p^{x^ + y"^) dy dx and only when 
p and ^ are constant can they be taken out from under the integral 
sign. If p and ^ are variable we have 






Pk {x^ + y^) dy dx. (3) 

(2) and (3) can be written in a form easily remembered, viz., 

I=Jr2dm, (4) 

where dm is an element of mass, and r is its distance from the 
axis. 

Sometimes in finding the moment of inertia of a body it is 
advantageous to choose the element of mass so that a single inte- 
gral will suffice. See for example Illustration 2, below. 

Illustration 1. Find / of a right-angled triangular plate whose 
thickness is 0.5 inch, and whose legs are 10 inches and 4 inches, 
about an axis through the vertex of the right angle and perpendicu- 



280 



CALCULUS 



[§144 



lar to the plane of the triangle. The density of the material is 
0.03 pound per cubic inch. See Fig. 105. 

I = 0.03-0.5 II (a;2 + y"^) dij dx. 



t/0 



The student will carry out the integration and find the radius of 
gyration. 

Illustration 2. Find 7 of a circular plate about an axis through 
its center and perpendicular to its plane. . The plate has a radius 
of 10 inches. It is 2 inches thick and its density is 0.04 pound 
per cubic inch. 

Hint. Here it is convenient to divide the plate into concen- 
tric rings of inner radius r and of width dr. See Fig. 106. The 





Fig. 105. 



Fig. 106. 



volume of such a ring is 2-2Trr-dr, and its mass is 0.04*47rr-dr. 
The distance of this mass from the axis is r. Hence 



I = 0.16 



■s: 



dr. 



Also find the radius of gyration. 

144. Transfer of Axes. Theorem. The moment of inertia of a 
body about any axis is equal to its moment of inertia about a parallel 
axis through the centroid, increased by the product of the jnass by 
the square of the distance between the axes. 

Let AB, Fig. 107, be the axis about which the moment of inertia 
is desired. Choose a system of rectangular axes such that the 
origin, 0, is at the centroid, such that the Z-axis is parallel to AB, 



§145] 



SUCCESSIVE INTEGRATION 



281 



and such that the ZF-plane contains the line AB. Consider an 
element of mass, dm, at P. The moment of inertia of the body 
about AB is then 

I = fiPBy dm = fUy - dy + x^l dm, 
or I = j {x"^ -\- y"^) dm — 2d j ij dm -{■ d^ j drn. (1) 

The first term of the right-hand side of (1) is the moment of 
inertia, /„, of the body about the Z-axis, an axis through the cen- 

troid. The second term, I y dm, is the moment of the body 

with respect to the XZ-plane, a plane passing through its centroid. 




Fig. 107. 



__ \y dm 



I dm 



Since y = 0, j y dm = 0. The last term, d"^ j dm, is d^M, 
where M is the mass of the body. Hence 

I = !„ + Md2. 

145. Moment of Inertia of an Area. We have spoken of the 
center of gravity of an area quite apart from any idea of mass and 
have stated that this is a useful conception in the study of mechan- 
ics. In the same way the solution of some problems in mechanics 
requires the moment of inertia of an area quite apart from any idea 
of mass. 



282 



CALCULUS 



[§145 



The moment of inertia of a plane area about an axis through the 
origin and perpendicular to its plane is defined by the integral. 

I = J J (x2 + y2) dy dx = Jf (x2 + y2) dx dy. 

The theorem on the transfer of axes holds in this case if the word 
"area" is substituted for the word "mass." 



Exercises 

Find the moment of inertia of the following : 

1. A rectangle of sides a and b about one comer. (See Fig. 104.) 
About the centroid. About one base. About a line parallel to one 
base and passing through the centroid. 

2. A right triangle, legs a and b, about 
one of the legs. About a line through the 

^^ centroid parallel to this leg. 

3. The area of a circle about an axis 
' through a point on its circumference and 

perpendicular to its plane. See Illustration 
2, §143. 

4. The area of a circle about a diameter. 
Hint. 




I = 



X? 



2xdy. 



Fig. 108. 



5. The area of a circle about a tangent 
line. 

6. The area between y = x^ and y^ = x 
about an axis through the origin perpen- 
dicular to the XF-plane. 

7. A uniform bar of length L and linear 
density p about an axis through one end 

perpendicular to the bar. Find / about a parallel axis through the 
middle point of the bar. 

8. A bar of length L, whose density is proportional to the distance 
from one end, about an axis perpendicular to the bar through the 
end of least density. 

9. A slender uniform rod. Fig. 108, about a line through its middle 
point and making an angle a with the rod. 

Ans. I — ^s'lnL^ sin'' a, where m is the mass and L is the length 
of the rod. 



§146] SUCCESSIVE INTEGRATION 283 

Hint. Denote by p the linear density. Then 

I = P i x^ sin^a dx, with proper limits. 

10. The rod of Exercise 9 about a parallel axis through one end. 

11. A wire bent into the form of a circular arc, Fig. 109, about the 
origin. Also find the moments of inertia, Ix and ly about the X- and 
F-axes, respectively. 



/- 
/- 



7 = I r^rdB; 
Ix = I r^sin^erdd', 



/„ = I r^ cos^e r de. 




Fig. 109. 



12. A triangle of base h and altitude h about an axis through the 
vertex parallel to the base. Divide the area into strips parallel to the 
base and of width dx. The axis of x is drawn from the vertex perpen- 
dicular to the base. 

_ , x^hx dx 



f 



13. A triangle about a line through the center of gravity parallel 
to the base. Use the result of Exercise 12. 

14. The area of the ellipse ^ + rj =1 about the major axis. (Use 

single integration.) About the minor axis. About the origin. 

146. Moment of Inertia: Polar Coordinates. The moment of 
inertia of the element r dr dd about an axis through the origin is 
r^r dr dd. Hence the moment of inertia of an area is 

/ = J Jr3 dr dd, 

with proper limits. If the moment of inertia of a plate is required, 
r' dr dd is to be multiplied by p, the density per unit area. 

Exercises 

Find the moment of inertia of the following : 

1. The area of the cardioid p = a (1 + cos 6) about an axis through 
the origin perpendicular to the plane of the cardioid. About the 
initial line. 



284 CALCULUS [§147 

2. The area of one loop of p = o cos 2d about the initial line. 

3. A circular sector of central angle 2a about the radius of 
symmetry. 

4. The arc of the sector of Exercise 3 about the radius of sym- 
metry. 

5. The area of Exercise 3, about an axis through the center of the 
circle and perpendicular to the plane of the sector. About a parallel 
axis through the centroid. 

147. Moment of Inertia of a Solid. — We wish to find the moment 
of inertia of the solid of Fig. 86, about the Z-axis. The moment of 
inertia of the element of mass, p dz dy dx, about the Z-axis is equal 
to pix"^ + 7/^) dz dy dx. The total moment of inertia of the solid 
about this axis is the integral of this element throughout the solid. 
Hence, 

/. = I I I p(x2 + 7/2) dz dy dx. (1) 



Similarly 



r, - f r r 

J a Jy^ Jz = 

n"2 ni = fix. 
J Jz = 

Jo, Jy, Jz = 



y) 
/«= I j I p{y^ + z^) dz dy dx, (2) 



•z - fix, y) 

p{z^ + x2) dz dy dx. (3) 



If the solid be regarded as a geometrical volume of density 1, 
the p's disappear, and the formulas (1), (2), and (3) can be written 

/. = Jffi^' + y') dz dy dx, (4) 

I. = Jffiy' + 2') dz dy dx, (5) 

ly = ///(2' + x^) dz dy dx. (6) 

Let 

/„, = III px^ dz dy dx, (7) 

Lx = JJJpy'^ dz dy dx, (8) 

hv = fffp^^ dz dy dx. (9) 

The quantities (7), (8), and (9) will be called the moments of 
inertia of the solid with respect to the FZ-plane, the XZ-plane, 
and the XF-plane, respectively. They are the integrals of the 



§147] SUCCESSIVE INTEGRATION 285 

product of the element of mass by the square of its distance from 
the respective planes. They can very frequently be found by a 
single integration by taking as element a plane lamina between 
two planes parallel to the plane with respect to which the moment 
is computed. If this is the case the moment of inertia about the 
coordinate axes can easily be found by noting that from the equa- 
tions(l), (2), (3); and (7), (8), (9): 



Iz 


= 


Iy^ 


+ Ixz, 


I. 


= 


Ixz 


+ Ixy, 


ly 


= 


Ixy 


+ Iyz. 



That is, the moment of inertia about the Z-axis is equal to the sum 
of the mom£nts of inertia with respect to the YZ- and XZ-planes, and 
so on. 

In general, the moment of inertia of a body about an axis is equxil 
to the sum of its moments of inertia with respect to two perpendicular 
planes which intersect in that axis. 

In the same way it follows from the formula for the moment of 
inertia of an area, I = I I (a;^ + y^) dy dx, that the sum of the 

moments of inertia of an area {or a plate) about two perpendicular 
axes is equal to its moment of inertia about an axis perpendicular 
to the plane of these axes through their point of intersection. 

Illustration 1. Find the moment of inertia of the ellipsoid 



X V z 

—i + T^-{ — s = 1 about each of its axes. 
a^ 0^ c'- 

First Method. 




y 



/. = 8j j ^ a^-^'\^ a^ %J^ + Z^) dz dy dx. 

Carry out the integration far enough to see that it is not simple 
and then note the relative simplicity of the 

Second Method. Compute Ixz, the moment of inertia with 
respect to the XZ-plane. Take as element of integration the ellip- 
tical plate cut out by the planes y = y and y = y -\- Ay. The 
equation of the intersection of the ellipsoid and the plane y = y is 
x^ 2^ _ 



.»(i-fi) .<!-«;) 



286 CALCULUS [§147 

Now the area of an ellipse is t times the product of its semi- 
major and semi-minor axes. Hence the area of this ellipse is 

The volume of the elliptical plate in question is 

irac [l - p-j dy 

and its moment of inertia with respect to the XZ-plane is 

xacy^ (l - p-j dy. 

The total moment of inertia of the ellipsoid with respect to this 
plane is then 

I,, = Tracfy^[l-pjdy 
= 27roc ( 



/63 5 A 



3 5/ 16 

I XV can be written down at once as 

-. _ 47ra6c' 

Then 

T TIT ^Trahc ,, „ . 

We can write down at once by interchanging letters: 

Illustration 2. Find the moment of inertia of a right circular 
cone about a line through its vertex perpendicular to its axis, if 
the radius of the base is b and the altitude is h. Choose the vertex 
as origin and the axis of the cone as axis of x. Consider the plate 

bx 
of radius r cut out by the planes x = x and x = x + dx. Its 



§147] SUCCESSIVE INTEGRATION 287 

moment of inertia about a diameter parallel to the axis of rotation 

through the vertex is equal to — j-rz — . (See expression for / of 

a circular plate about a diameter, Exercise 4, §145). Then the 
moment of inertia of this plate about the axis of rotation through 
the vertex is equal to this moment of inertia about an axis through 
the centroid increased by its volume (mass if density = 1) multi- 
plied by the square of the distance between this axis and the par- 
allel axis through the vertex (see §144) i.e., to 

Tb*x*dx irb^x'^dx ^ 
W "^ P ^ ■ 

And the total moment of inertia of the cone about the axis through 
the vertex is equal to the integral of this moment of inertia from 
X = to X = h. That is 



SX 






■jrb^h , Trb%^ Trb-h ,, „ , ^,,. 
"20- +-5- = -20" (^^ + ^'^^)- 



Exercises 



Find the moment of inertia of: 

1. The cone of Illustration 2, about a parallel axis through the 
centroid of the cone. About a diameter of the base. 

2. A right circular cylinder, the radius of whose base is r, and whose 
altitude is h, about a diameter of one base. About a parallel axis 
through the centroid. 

3. A rectangular parallelopiped with edges a, b, and c, about an 
axis through the centroid parallel to one edge. 

4. A right circular cylinder about its axis. 

5. A hollow right circular cylinder of outer radius R, inner radius r, 
and altitude h, about its central axis. About a diameter of one base. 
About a diameter of the plane section through the centroid perpen- 
dicular to the axis. 

6. A right rectangular pyramid of base a X b and of altitude h, 
about an axis through the centroid parallel to the edge a. About an 
axis through the vertex and the center of gravity. 

Ans. /. = ^ (b- + Ih^). h=^ {a' + &=). 



288 CALCULUS [§147 

7. A frustum of a right cone about its axis if the radius of the large 
base is R, that of the small base is r, and the altitude is h. 

R^ — r^ 

Ans. I = -fifTrh -fz 

R — r 

8. A hollow sphere about a diameter, if the outer radius is R and 
the inner radius is r. 

9. A paraboloid of revolution, the radius of whose base is r and 
whose height is h, about the axis of revolution. 

Ans. I = lirr*h. 

10. The anchor ring generated by revolving the circle 
[x - R]^ + y^ = r^ about the F-axis. 

Ans. h = TT^RrKR^ + fr"). ly = 27rmr\R^ + fr^). 

11. A right circular cone about its axis. 

12. A right elliptical cylinder of height L, and having the semi-major 
and semi-minor axes of its cross section equal to a and b, respectively, 
about an axis through the centroid parallel to b. 

13. A quadrant of a circular plate about one of its bounding radii. 

14. An equilateral triangle of side 2a, about a median. About a 
line through a vertex perpendicular to one of the sides through that 
vertex. 



CHAPTER XVI 



CURVATURE. EVOLUTES. ENVELOPES 



y 




1 

qI 




A- 


Pf /r+t^T 


o 


y 


/ 



Fig. 110. 



148. Curvature. Let PT and QT, Fig. 110, be tangents drawn 
to the curve APQ at the points P and Q, respectively. Denote the 
length of the arc PQ by As and 
the angles of inclination of PT 
and QT' to the positive direc- 
tion of the X-axis by t and t 
-j- Ar, respectively. At gives a 
rough measure of the deviation 
from a straight line, of that por- 
tion of the arc of the curve be- 
tween the points P and Q. The 
sharper the bending of the curve 
between the points P and Q the 

greater is At for equal values of As. The average curvature of 
the curve between the points P and Q is defined by the equation 

At 
Average Curvature = 't~'' (1) 

The average curvature of a curve between two points P and Q 
is the average change between these points, per unit length of 
arc, of the inclination to the X-axis of the tangent line to the 
curve. Or, more briefly, the average curvature is the average 
change per unit length of arc, in the inclination of the tangent line. 

The curvature at P is defined as the limit of the average curvature 
between the points Q and P as Q approaches P. On denoting the 
curvature by K, we have in accordance with the definition, 

^' ^' (2) 



K = 1™ 

■^ As = As 



ds 



The curvature at a point P is then the rate of change at this point of 
the inclination of the tangent line per unit length of arc. The 
19 289 



290 



CALCITLUS 



I §151 



curvature is a measure of the amount of bending of a curve in 
the vicinity of a point, 

149. Curvature of a Circle. It is clear that the average curva- 
ture of a circle, Fig. Ill, is 



At _ At 
As r At 



Hence the average curvature is independent of As and conse- 
quently the curvature, the limit of the average curvature as 
As approaches zero, is 



A' = 



(1) 



Tfie curvature of a circle is constant and equal to the reciprocal of its 

radius. 

150. Circle of Curvature. Radius of Curvature. Center of 

Curvature. Through any point P of a curve infinitely many cir- 
cles can be drawn which have a 
common tangent wdth the curve at 
P and whose centers are on the con- 
cave side of the curve. Of these 
circles there is one whose curvature 
is equal to that of the curve at P, 
i.e., one whose radius is equal to the 
reciprocal of the curvature at P. 
This circle is called the circle of 
curvature at the point P. The 
radius of this circle is called the 

radius of curvature, and its center the center of curvature, of the 

curve at the point P. The radius of curvature is denoted by R 

and, in accordance with (2), §148, its length is 




Fig. 111. 



R = 



K 



ds 
dr* 



(1) 



151. Formulas for Curvature and Radius of Curvature: Rec- 
tangular Coordinates. For obtaining the curvature at any point 
on the curve y = /(x), we shall now develop a formula involving 



§151] CURVATURE. EVOLUTES. ENVELOPES 291 

the first and second derivatives of y with respect to x. The above 
formula for curvature K can be written 



(1) 





dr dr 

dx dx 




dx \1 + [dx) 


Since 


. ,dy 

T = tan-^ -j-y 

dx 




d'y 
dr dx"^ 


Consequently 




and, by (1), §150, 


[^M^Q? 




, [-(S)r 

^ - d2y 
dx2 



(2) 



(3) 



We shall understand by K and R the numerical values of the 
right-hand members of (2) and (3), respectively, since we shall 
not be concerned with the algebraic signs of K and R. 

Illustration. Find the curvature oi y = x^. 

^ = 2x ^ = 2 
dx ' dx^ 

Substitution in formula (2) gives 

2 



K = 



(1 + 4x2)^ 



From this expression it is seen that the maximum curvature 
occurs when x is zero, and that the curvature decreases as x 



292 CALCULUS [§152 

increases in numerical value. When a; = 0, K = 2. When 



> — X -1-, -IV — 


25 


Exercises 




Find the curvature and radius of curvature of each of the curves: 


l.y = 2x- xK 




A. y = x^ — x^. 


7. 2/ - 3x*. 


2. 2/ = xi 




5. y = -.• 


8. y = x~^. 


-y-l- 




6. 2/ = Vx. 


Vx 



10. If p = /(&) is the equation of a curve in polar coordinates, show 
that 

„ ^ +Hd"gJ -^d¥^ 
K — , • 



Hint. 



['-(^^)T 



_ dr _ d^ 

"~ ds ~ ds 

d« 

T = e +ip. (See Fig. 72.) 

til- _ d^ 

50 ~ "^ do" 



Obtain -jr from the relation 
dff 



d0 



IS given in 



de 

152. Curvature: Parametric Equations. If the equation of a 
curve is expressed in parametric form, z = f{t), y = F{i), the cur- 
vature can be found by differentiating x and y and substituting 
in (2), §151. t can be eliminated from the result if desired. 

Illustration 1 . li x = t and y = t^, 

^ ^ 2« and ^ = ^^ — = 2 
dx ' dx' dt dx 



§153] CURVATURE, EVOLUTES. ENVELOPES 293 

Hence 

2 2 



Illustration 2. Find the curvature of the ellipse x = a cos t, 
y = 6 sin t. 



dy b 

-J- = cot t. 

ax a 



d-hj b d .dt & „, 

3-: = -V, cot t -3— = — csc^ t 

dx^ a at ax a 



r— T = iCSC^^ 

a sin t_\ a^ 



- -cscH 
„ a^ — ab — ab 

h. = 



[l+^'cot^f]^ (a'sinH + b^cosH)^ g ^2 + ^J ^^1 



a*b* 



ib*x^ + a*y^Y 

Exercises 

1. Find the curvature of the curve x = a cosh I, y — a sinh t. 

2. Find the curvature of the curve x = a{t — sin 0) 
y = a{\ — cos 0- 

153. Approximate Formula for the Curvature. If a curve 

dy 
deviates but little from a horizontal straight line, -r- is small and 

(dv\ 2 
^1 in formula (2), §151, is very small compared with 1. 

Hence the denominator differs very little from 1 and the formula 
becomes approximately 

^=S- (') 

This approximate formula for K is frequently used in mechanics 
in the study of the flexure of beams. The slope of the elastic 

d'^y 
curve of a beam is so small that -v-^ can be used for the curva- 
ture without appreciable error. 

The approximate formula for the radius of curvature R is 

7?- ^ 

^~ d^ (2) 

dx^ 



294 



CALCULUS 



[§154 



154. Center of Curvature. Evolute. Formulas will now be 
obtained for the coordinates of the center of curvature of a curve 
corresponding to any point P. Let the coordinates of P be a; and 
y. Denote by a and j8 the coordinates of the center of curvature 
of the curve at this point. There are four cases to be considered. 
See Fig. 112, a, b, c, d. 




Fig. 112. 



In Fig. 112, a, 



a = OM =^ ON - HP = X - Rsinr, 
/3 = MC = NP + HC = y + R cost. 



Since 



tanr = 



dy 
dx 



cost = 



NRS-f 



; sinT 



dy 

dx 



/dy\^ 
\dx) 



§154] CURVATURE. EVOLUTES. ENVELOPES 295 

Consequently, 



^-m 



«=--f-d^f^' (1) 



and 



1 + 



(^) 



^ = y + — d^- (2) 

The student can show that, since -7— is negative for a descending 

curve and positive for an ascending curve, and since -r^ is positive 

when a curve is concave upward and negative when a curve is 
concave downward, formulas (1) and (2) hold for the three curves 
represented in Fig. 112, b, c, d. 

Illustration. Find the coordinates of the center of curvature 
corresponding to any point on the curve y = ± 2 y/~x. Only the 
positive sign will be used. If the negative sign is used it will 
only be necessary to change the sign of ^. 

dy ^ 1 

dx y/^ 

^ = _ J_ 

dx^ 2a;i* 

1 



1 



1 + 



2x1 



= y 1 — = y 



- 2 V^(x + 1) = 2/ - 2/ [I + 1] = - 



2x3 

The equation of the locus of the center of curvature is obtained 

by eliminating x and y from the equations for a and /3 and the 

equation of the original curve. Thus 

0= ~ 2 /.ox' 

X = — y— ; y = - (4)8)' . 



296 CALCULUS [§155 

Substituting in y"^ = 4a;, we obtain 

^^ = A(« - 2)», 
the equation of the locus of the center of curvature. This is the 
equation of a semi-cubical parabola whose vertex is at the point 
(2,0). 

The locus of the center of curvature corresponding to points on a 
curve is called the evolute of that curve. Its equation is easily 
obtained in many cases by eliminating x and y from equations (1) 
and (2) and the equation of the original curve. Otherwise (1) 
and (2) constitute its parametric equations, a and /3 being ex- 
pressed in terms of the parameters x and y. 

Exercises 

1. Find the evolute of t/ = 4x^. 

2. Find the evolute of the ellipse 



I* 



^ ^yi - 1 

a^ "^ 6* ^' 

Hint It will be found that 

(a* - &2)x' (a« - 6*)y« 
« = ^, ; ^ = ^, 

Whence 

Elimination gives 

iaa)i + (6^)5 = (o2 - 62)5. 

3. Find the parametric equations of the evolute of the cycloid, 

X = a{9 — sin 9), 
y = a(l — cos e). 
Ans. a = a(0 + sin 6), = — a(l — cos d). Show that the evo- 
lute is an equal cycloid with its cusp at the point ( — wa, — 2a). 

4. Find the equation of the evolute of 

X = a(cos 9 + 9sm e), 
y = o(sin — ^cos^). 
Ans. a = a cos 9, /3 = a sin 9. Discuss. 

155. Envelopes. If the equation of a curve contains a constant 
c, infinitely many curves can be obtained by assigning different 
values to c. Thus 

(x - c)2 + 2/2 = a2 (1) 



§155] 



CURVATURE. EVOLUTES. ENVELOPES 



297 



is the equation of a circle of radius a whose center is at (c, 0). By- 
assigning different values to c we get a system of equal circles 
whose centers lie on the X-axis. A constant such as c, to which 
infinitely many values are assigned, is called a parameter. A 
constant such as a, which is thought of as taking on only one value 
during the whole discussion, is called an absolute constant. We say 
that equation (1) represents Si family of circles or a system of circles 
corresponding to the parameter c. 

The general equation of a family of curves depending upon a 
single parameter can be written in the form, 



fix, y, c) = 0. 



(2) 



Exercises 
State the family of curves represented by the following equations 



contammg a parameter: 

1. y = x^ + c, 

2. y = mx + b, 

3. y = mx + b, 

*• a» ^ &» ^' 
6. 2/* = m(x + m), 
6. x^ + y^ = a\ 



c being 
b being 
m being 

a being 

TO being 
a being 



the parameter, 
the parameter, 
the parameter. 

the parameter. 

the parameter, 
the parameter. 



Consider again the family of circles (1). Two circles of the 
family corresponding to the values, c and c -f- Ac, of the parameter 
intersect in the points Q and Q', 
Fig. 113. We seek the limiting 
positions of these points of in- 
tersection as Ac approaches zero. 
Clearly they are the points P 
and P', respectively, on the lines 
y = +a. Such a limiting posi- 
tion of the point of intersection 
of two circles of the family is 
called the point of intersection of 
two "consecutive" circles of the 

family. In general, the limiting position! f the point of intersec- 
tion of two curves, f{x, y, c), f (x, y,c-\- Ac), of a family, as Ac 




298 CALCULUS [§155 

approaches zero, is called the point of intersection of "consecu- 
tive" curves of the family. 

In the case of the family of circles (1) the locus of the points of 
intersection of "consecutive" circles is the pair of straight lines 
y = +a. This locus is called the envelope of the family of circles. 
In general, the envelope of a family of curves depending upon one 
parameter is the locus of the points of intersection of "consecutive" 
curves of the family. It will be shown in a later chapter that the 
envelope of a family of curves is tangent to every curve of the 
family. 

Exercise 

Draw a number of lines of the family 

X cos 6 + y sin d = p, 

where 9 is the parameter, and sketch the envelope. 

A general method of obtaining the envelope of a family of 
curves will now be given. 

The equation of a curve of the family is 

f(x, y, c) = 0, (3) 

where c has any fixed value. The envelope is the locus of the 
limiting position of the point of intersection of any curve (3) of 
the family with a neighboring curve, such as 

fix, y, c + Ac) = 0, (4) 

as the second curve is made to approach the first by letting Ac 
approach zero. The coordinates of the points of intersection of 
the curves representing equations (3) and (4) satisfy 

fix, y, c + Ac) - fix, y, c) = 0. (5) 

Then they satisfy 

fix,y,c + Ac) -fix,y,c) 



Ac 



= 0, (6) 



since Ac does not depend on either x or y. Then the coordinates 
of the limiting positions of these points of intersection satisfy 

lim fix,y,c + Ac) -fix,y,c) _ 
Ac A ^c ~ 



§155] CURVATURE. EVOLUTES. ENVELOPES 299 

The first member of this equation is the derivative of /(x, y, c) 
with respect to c. It may be written in the form, 

df(x, y, c) _ , 

dc ~ ^- ^'' 

The differentiation is performed with respect to c, x and y being 
treated as constants. The point of intersection also lies on (3). 
Hence the equation of its locus is obtained by eliminating c 
between (3) and (7). 

Illustration 1. Find the equation of the envelope of the 
family of circles, {x — cy -{- y^ = a^, c being the parameter. 

The equation of the curve written in the form f{x, y, c) = is 

{x - cY + 2/2 _ a2 = 0. (I) 

Differentiating with respect to c, 

- 2{x - c) = 0. (II) 

The elimination of c between (I) and (II) gives 

or 

y = ±a, 
as the envelope. 

Illustration 2. Find the equation of the envelope of the 
family of lines, x cos 6 -^ y sin d = p, 6 being the parameter. 
On differentiating the first member of 

X cos d -\- y sin d '- p = (I) 

with respect to d we obtain 

— X sin 6 -\- y cos d = Q. (II) 

The result of eliminating d between (I) and (II) is 

^2 _j_ ^2 _ p2^ 

a circle of radius p about the origin as center. 

Exercises 

P 
1. Find the envelope of the family of straight lines y — mx -\ — ' 

m 

where m is the parameter. Draw figure. 



300 CALCULUS [§156 



2. Find the envelope of the family of lines y = mx + o\/l + m', 
where m is the parameter. Draw figure. 

3. Find the envelope of the family of parabolas y^ = c(x — c), c 
being the parameter. 

4. Find the envelope of the family of lines of constant length whose 
extremities lie in two perpendicular lines. 

6. Find the envelope of y = px — -p^, p being the parameter. 
Draw figure. 

6. Find the envelope of the family of curves (x — c)^ + 2/^ = 4pc, 
c being the parameter. Draw figure. 

7. The equation of the path of a projectile fired with an initial 
velocity Vo which makes an angle a with the horizontal, is 

y — X tan a — n — ; 7. — 

" 2v^ cos^a 

Find the envelope of the family of paths obtained by considering a a 

parameter. 

Wo* gx^ 
Ans. V == -ji— — 77—! 

8. The equation of the normal to y^ = 4x at the point P, whose 
coordinates are Xi and yi, is 

yi , ■. 

y - yi = - 2^ (x— xi). 

Since yi^ = 4xi, this may be written 

2/iX +2y -'^ 2yi= 0. 

Find the equation of the envelope oi the normals as P moves along 
the curve. 

Hint. On differentiating with respect to the parameter 2/1 we 
obtain 

a. o V^^^Z? 

On substituting this value of yi in the equation of the normal and 
squaring we obtain 

2 4(x - 2)^ 

y = —27 — 

This is the evolute of the parabola as we have seen in §164. 

156. The Evolute as the Envelope of the Normals. In Exercise 
8 of §155 it was seen that the evolute of a parabola is the envelope 
of its normals. This is true for any curve. The result is fairly 
evident from the examination of the curves of the exercises of 



§156] CURVATURE. EVOLUTES. ENVELOPES 301 

§165 and their evolutes. It will be shown that the normals to a 
curve are tangent to its evolute. 

The parametric equations of the evolute are 

a = X — Rsinr, (1) 

^ = y + RcosT. (2) 

On differentiating with respect to the variable s, which is per- 
missible since x, y, R, and r are aU functions of s, we obtain 

da dx dR . _, dr 

3— = 3 5— sm T — /t cos r 3- • 

ds ds ds ds 



d^ dy , dR „ . dr 

T" = 3 r T' COST — R smr -j-- 

ds ds ds ds 



Now 



= COST, 



dx 
ds 
dy 

ds ~ R' 

Then the foregoing equations become 
da dR . 

ds ds ' 

d^ dR 

3— = 3— COS T. 

ds ds 
Hence the slope of the tangent to the evolute is 

^ = -cotT (3) 

da 

Therefore the tangent to the evolute is parallel to the normal to 
the curve at the point (x, y) to which (a, jS) corresponds. But the 
normal to the curve at (x, y) passes through (a, /3). Hence it is 
tangent to the evolute at this point. 

It can also be shown that if Ci and C2, Fig. 114, are the centers 
of curvature corresponding to the points Pi and P2, the length of 
the arc C1C2 of the evolute is equal to the difference in the lengths 
of the radii of curvature, Ri and R^. For, from the above values 
of da and d^ it follows that 

Vda^ + diS* = dR. 



302 CALCULUS [§157 



But VdoM-d^ is the differential of the arc of the evolute. Call 
it d(T. Then da = dR, and hence on integrating a = R -\- C. 
a and R are functions of s, the arc of the given curve. Then corre- 
sponding to a change As( = arc P1P2) in s, cr and i2 will take on the 
increments Ac and AR which are equal by the foregoing equation. 
But A(T = arc C1C2, and AR = R2 — Ri. Hence arc C1C2 equals 
^2 — Ri- 




Fig. 114. 

157. Involutes. In Fig. 114, suppose that one end of a string is 
fastened at C and that it is stretched along the curve CdCiKM. 
If now the string be unwound, always being kept taut, the point M 
will, in accordance with the properties of the evolute, trace out the 
curve MP1P2P. This curve is called the involute of the curve 
KC1C2C. If longer or shorter lengths of string, such as CKM2 
be used, other involutes will be traced. In fact to a given curve 
there correspond infinitely many involutes. The given curve is 
the evolute of each of these involutes. We see that while a given 
curve has but one evolute it has infinitely many involutes. 

In Exercise 4, §154, the circle x = a cos 6, y = asin 6 was found 
as the evolute of the curve x = a(cos 6 -\- 6 sin 6), y = a(sin 6 — 
6 cos 6). Then the latter curve is an involute of the circle. The 
student will draw a figure showing a position of the string as it 
would be unwound to generate the involute and indicate the 
angle 6. 



CHAPTER XVII 

SERIES. TAYLOR'S AND MACLAURIN'S THEOREMS. 
INDETERMINATE FORMS. 

158. Infinite Series. The expression 

Ui + U2-\- U3-\- ■ ■ ■ + Un+ ■ ' ' (1) 

where ui, Uz, "Wa, • • •, %i„, ' • ' is an unlimited succession of 
numbers, is called an infinite series. 

Let s„ denote the sum of the first n terms of the infinite series 
(1). Thus 

Sn = Ul + U2 + U3 + • • • -\- u„. (2) 

If, as n increases without limit, Sn approaches a limit s, this 
limit is called the sum of the infinite series, and the series is said 
to be convergent. 

Illustration 1, In Illustration (1), §21, AB, Fig. 21, is a line 
2 units long. The lengths Axi, X1X2, XiXz, x^Xt, • • • x„_ix„, 

• • • are 'i-, h, l, h ' ' ' > n~i t ' ' ' > respectively. For this 
series 

Sn = 1 + ^ + ^ + I + • • • + -^f (3) 

The limit of this sum as n increases without limit is 2, as the 
figure shows. Or, we may write, 

l + h + l + l+ • • • +2^,+ • ■ ' =2. (4) 

Illustration 2. The sum of the geometrical progression 

1 + r + r^ + r^ + r" + • • • + r" (5) 

does not approach a limit if \r\ y 1, but if \r\ < 1 it approaches 

the limit z , when n becomes infinite. 

1 — r' 

The infinite series (1) is said to be divergent or to diverge if, as n 

increases without limit, s„ does not approach a limit. 

303 



304 CALCULUS [§158 

Thus the series 

l+2 + 3 + 4+---+n + ---, 

l-1 + l-l + l-l + l---, • 

1 + R-\- R^-\- R^ + R*+ • ■ • + R^+ • ■ ' {R> I) 

are illustrations of divergent series. 

If the terms of an infinite series are functions of a variable x and 
if the series is convergent for any range of values for x, the series 
defines a function of x for that range of values. Thus, if |x| < 1, 
the series 

1 +x + x^ + x^ + • • • +x^ + • ' ' (7) 

defines the function z . On the other hand, if the series is 

1 —x ' 

divergent it does not define a function of x. Thus, if \x\ > 1, the 

series (7) is divergent and does not define the function ^ _ ' 

or any other function. 

It may happen that the sum of a few terms of an infinite series 
representing a function is a very close approximation to the value 
of the function. As an illustration take the infinite geometrical 

progression (7), which when |x| < 1 represents the function . _ . 

If the terms after x*~^ are neglected, the error is 

X* + x*+i + • • . + X" + • • • = r^—- 

1 — X 

X* 

The error, :j-^ — is very small compared with the value of the 

function, q , and decreases as k increases; i.e., a better and 

1 — X ' ' 

better approximation is obtained the greater the number of terms 

retained. 

Another infinite series is obtained by expanding (1 + x)' by 
the binomial theorem, 

(1 + x)' = 1 + -|x + Ix' - -Ax^ + • • • (8) 

This series can be shown to be convergent when |x| < 1 and 
divergent when |x| > 1. 

Just as the function . _ is represented to a high degree of ap- 
proximation by the first few terms of the series (7) when jxj is small, 



§159] TAYLOR'S AND MACLAURIN'S THEOREMS 305 

the function {I ■}- x)^ is represented approximately by the first 
few terms of (8) when |a;| is small. In both cases the functions are 
represented approximately by polynomials. A method will be 
developed in the succeeding articles which will enable us to deter- 
mine polynomial approximations to other functions, such as sin x, 
tan X, e'. 
An infinite series of the form 

ao+ di'X + o.-ix''' + asx' + . . . + OnX" + . . . (9) 
is called a 'power series in x. One of the form 
Oo + ai(a; — a) + a^ix — ay + 03(3; — a)' + • • • 

+ an{x - a)" + • • • (10) 

is called a power series in (x — a). The series (7) and (8) are 
power series in x representing the functions ^ _ and (1 + x)^, 

respectively. In the succeeding articles power series will be 
obtained representing the functions sin x, tan x, e*, etc. 

159. Rolle's Theorem. Let f{x) be a single-valued continuous 
function between x = a and x = b, having a continuous first 





Fig. 116. 




Fig. 117. 



Fig. 118. 



derivative, f'{x), between the same limits. Further, let /(o) = 
and f(b) = 0, i.e., let the curve representing the function cross or 
touch the X-axis at a; = a and x = b. The curve may or may not 



?o 



306 



CALCULUS 



[§160 



cross or touch the X-axis at intermediate points. (See Fig. 115.) 
Since f{x) is continuous it cannot have a vertical asymptote be- 
tween X = a and x = b as shown in Fig. 116, nor can it have a 
finite discontinuity as shown in Fig. 117. Since f'{x) is con- 
tinuous between x = a and x = h, the curve cannot change its 
direction abruptly between these limits, as shown in Figs. 118 and 
119. Since cases such as are represented by Figs. 116, 117, 118, 
and 119 are excluded, the curve. Fig. 115, must have a horizontal 
tangent at some point x = Xi between x = a and x = b. Hence 
the 





Fig. 119. 



Fig. 120. 



Theorem. If f{x) is a single-valued function from a; = a to 
X = b, andiff(x) andf'(x) are continuous between these limits, and 
further if f{a) = andf(b) = 0, thenf'ixi) = 0, where a < Xi < b. 

160. Law of the Mean. Let f(x) be a single-valued function 
between x = a and x = b. Further let f{x) and f'{x) be con- 
tinuous between these limits. Fig. 120. It is then apparent from 
the figure that at some point P between A and B, the tangent 
line to the curve will be parallel to the secant line AB. Hence the 

Theorem. If f{x) and f'{x) are continuous between x = a and 
X = b, then 

J ^""'^ - b-a ' 
where a < Xi < b, or 

fib) = /(a) + (& - a)f(x^). (1) 

An analytic proof of this law will also be given. Define a 
number Si by the equation 
fib) = fia) + ib- a)Si, or 

fib) -fia) -(6-a)>Si = 0. (2) 



§161] TAYLOR'S AND MACLAURIN'S THEOREMS 307 

It will be shown that Si = f'{xi), where a < Xi < b. From the 

first member of (2) build up the function 0i(a;) by replacing a 
by X. 

<l>iix) =m -fix) - {b-x)Su (3) 
Then 

<f>'i{x) = -f{x)+Su (4) 

Since f{x) and f'(x) are continuous between x = a and x = b, 
0i(a;) and <f>'i(x) are continuous between the same limits. By (3) 
and (2), <^i(a) = 0, and by (3), <t>i{b) = 0. Hence <^i(x) satisfies 
the conditions of RoUe's Theorem and consequently 

<l>'i(xi) = 0, 
or 

/'(xi) - 5i = 0, 
or 

Si^f'ixi), 

where a < xi < b. On substituting this value of Si in (2) we 
obtain 

m =/(a) + (6-a)/'(xi), 

which proves the theorem. 

161. The Extended Law of the Mean. Let f(x) be a function 
which with its first and second derivatives, f'{x) and /"(x), is 
continuous from x = a to x ^ b. Define a number Sz by the 
equation 

m = f{a) + {b- a)f'{a) + ^-^-^S^, (1) 

or 

m - /(a) - (6 - a)f'ia) - ^^^^S, = 0. 

From the first member of the latter equation, form the func- 
tion (f>2{x) by replacing a by x: 

Mx) = m - /(x) - (6 - x)/'(x) - ^^^'-5,. (2) 

Then 

0'2(x) = -fix) - (6 - .x)/"(x) +/'(x) + (& - x)S, 
= {b-x) \S^ - /"(x)]. 

Since /(x), /'(x), and/"(x) are continuous, ^lix) and 4>\{x) are 
continuous. Further by (2) and (1), 02(a) = 0, and by (2), 



308 CALCULUS [§162 

02(&) = 0. Hence the conditions of RoUe's Theorem are satisfied, 
and 

<t>'2{x,) = 0, (3) 

where a < X2 < h. 
Or 

[h - X.MS2 - fix,)] = 0, 
or 

^2=r(x2). (4) 

On substituting this value of Sz in equation (1), we obtain 

m = /(a) + (6 - a)na) + ^^^^fix,), (5) 

where a < x^ <. h. 

162. Taylor's Theorem with the Remainder. Finally let j{x) 
and its first n derivatives be continuous from x = a to x =b. 
Define Sn by the equation 

m = f{a) + (6 - a)r{a) + ^^^^f{a) + ■ • • 



m - m -ib- a)f'{a) - ^^-—^fia) - • • • 

I n — 1 "^ I n 

Form the function </)„(x) by replacing a in the last equation by x. 

*.(x) =./(6) -/(x) - (6 - i)/'(x) - ^^-=^V(^) - ■ • ■ 

_((^)^ (6-x)« 

n — 1"^ _ri 

Then 

<^ "C^) = - |n_i /^"^(^) + |^_i 'S:,. (3) 

Since /(x),/'(x), • • •, /W(x) are continuous, <^„(x) and </)'„(a;) 
are continuous. By (2) and (1) <i)n{a) = 0, and by (2) 4>n{h) = 0. 
Hence the conditions of Rolle's Theorem are satisfied, and 

'P'niXn) = 0, 



§162] TAYLOR'S AND MACLAURIN'S THEOREMS 309 
or 

S„=f(-^{Xn) (4) 

where a < x„ < 6. Hence the 

Theorem: If f{x), f{x), f"{x), • • •, /("Ka;) are continuous 
from X = a to X = b, 

m = f(a) + (b- a)!' (a) + ^^^^ r(a) + ■ ■ ■ 

where a <. x„ <. b. 

This theorem, which is only an extension of the theorem express- 
ing the law of the mean, is called Tmjlor's Theorem with the 
remainder. The last term is called the remainder. 

If b is replaced by x, (5) becomes 

/(x) = /(a) + (X - a)}\a.) + ^"^'/'(o) + • ■ - 

where a < x„ < x. This inequality is sometimes written 
Xn = a -\- e{x - a), where < B < I. 
Illustration 1. Let/(x) = e*. Then 

/(x) = e' f{a) = e» 

fix) = e» /'(o) = e» 

/"(x) = e' /"(a) = e- 

/(")(a;) = e* f^"\a) = e- 



Hence by (6) 



l + (x-a)+-^"-^+ • • • 4-(^-«)"-^ 



In - 1 



+ (^^^,... (7) 



If a = 0, 

If a = and x = 1, 



«^2 /j«7l ~" 1 ']*'* 



e=l + l + ,^+. • •+r^ + T^«-- (9) 



310 CALCULUS [§162 

The remainder in (8) and (9) can be made as small as we please 
by choosing n sufficiently large. 

Taylor's Theorem may be expressed in still another form by 
setting b in (5) equal to a -\- h. 

fia + h)= Si.a) + hna) + ^ /"(a) + • • • + ^/^"^(xn) (10) 

where a < x„ < a + A, or Xn = a + 0/i, < ^ < 1. 

If the values of a function and its derivatives are known at a, 
then the values of the function at a point a -\- h can be computed 
by this formula. 

In (10), /(a + A) is represented approximately by a poly- 
nomial of degree n — \ in h. The coefficients are the derivatives 
of /(x) at X = a. The error in the approximation is given by 
the last term. This term gives only a means of estimating the 
error, since Xn is not known. The maximum error can, however, 
be determined by substituting M^^\ the greatest numerical 
value of /^">(a;) in the interval (a, a -\- h) for f''\xr.). The 
numerical value of the error is therefore less than 

|n 
If a = 0, (6) becomes 

f(x) = f(0) +f (0) X + f"(0) % + ' • • + f^"~'' (0) ^5^^ 

12 l(n-l) 

+ f^"nx»)^» (11) 

where < Xn <• x, ov Xn = 6x, < d < 1. 

In (11) it is assumed that the function f{x) and its first n 
derivatives are continuous from x = to x = a. (11) is known 
as Maclaurin's Theorem with the remainder. 

Illustration 2. Expand sin x by Maclaurin's Theorem in powers 
of z as far as the term containing x®. 



/(x) = sin X 


/(O) 


= 





f{x) = cos X 


/'(O) 


= 


1 


/"(x) = - sin X 


/"(O) 


= 





fix) = - cos X 


/'"(O) 


= 


- 1 


/■^(x) = sin X 


rxo) 


= 





P{x) = cos X 


/"(O) 


= 


1 


/^*(x) = — sin X 


rm 


= 





/^"(x) = - cos X 


r'\x.) 


= 


— COS X7. 



§163] TAYLOR'S AND MACLAURIN'S THEOREMS 311 
Substitution in (11) gives 



X'' 



sina; = x — ro+j-g— i^ cos Xi, (12) 



where <. Xt <. x. 

Since |cos X7I < 1, sin x differs from 
x^ x^ 

^~ [3 + j5 

x'' 
by a number less than r«* 

In general, since sin x and its derivatives are continuous, 

x^ , x^ x" , , ^ , s 

smx = X — r^ + re ~ ' ' "if" (sin x„ or cos x,,), (13) 

where < x„ <. x. Thus the difference between sin x and 

^ ~ [3 + [5 ~ ■ ■ " - \r^^^ 

x" 
is less than r—, a number which for a given x can be made as small 
\n 

as we please by taking n sufficiently large. Hence the series (13) 
can be used in computing the value of sin x. If x is small, only a 
few terms of the series need be used to obtain a very close ap- 
proximation to sin X. Thus in formulas in which sin x occurs, 
sin X is frequently replaced by x if the angle is small. Such a sub- 
stitution was made in equation 1, §81. It must be remembered 
in making the substitution that x is expressed in radians. 

163. Taylor's and Maclaurin's Series. If f{x) and all of its 
derivatives are continuous within an interval, the number of 
terms in (6), (10), and (11), §162, can be increased indefinitely. 
These equations then become, respectively, 

f (x) = f(a) + f (a) (X - a) + f"(a) ^^^' + • • • 

+ f^"Ha)^^^%---. (1) 

f(a+h)=f(a) + f'(a)h + r(a)j'^+- • • + f (">(a)^+ • • •. (2) 

f(x)=f(0) +f (0)x+ f"(0)^^ + • • • + f ^"^ (0)^ + • • •• (3) 



312 CALCULUS [§163 

In (1), f{x) and its derivatives are assumed to be continuous 
from a to x. 

In (2) , /(x) and its derivatives are assumed to be continuous from 
aio a + h. 

In (3), /(x) and its derivatives are assumed to be continuous 
from to X. 

The series (1) and (2) are called Taylor's Series and (3) is called 
Maclaurin's Series. 

If we denote the last term in each of the equations (6), (10), and 
(11), §162, by Rn, it is necessary that 

i^h^n = 

in order that (1), (2), and (3) shall represent /(x), /(o + h), and 
/(x), respectively. 

Such series represent a function only so long as they are conver- 
gent. Later in this chapter means of testing the convergence of 
series will be discussed. The series (1), (2), and (3), if convergent, 
represent /(x) but do not give a means of estimating the error 
made by stopping with a given term. This can best be deter- 
mined from the expression for the remainder 72„ in Taylor's or 
Maclaurin's Theorem with the remainder. 

Illustration 1. Represent sin x by a power series in (x — a). 
Use formula (1). 

/(x) = sin X /(a) = sin a 

/'(x) = cos X f'ia) = cos a 

j" (x) = — sin X /"(o) = — sin a. 

/'"(x) = — cos X /'"(«) = — cos a 

fix) = sin X /'^(«) = sin a 

P{x) = cos X /^(o) = cos a 

Then by (1) 

, , . . (x- ay (x - ay 

sin X = sin a + cos a {x — n) — sm a — r^ cos a — r^ — 

. (x - a)* {x -ay 

+ sm a — Hi + cos a , - — - — • • •. 

The corresponding Maclaurin's Series is obtained by letting a = 0. 

x' , x^ x^ , 
sin X = x - rg + , 5 - 1^ + • . •. 



§164] TAYLOR'S AND MACLAURIN'S THEOREMS 313 

Illustration 2. Expand tan x in a power series in x. 

fix) = tan X. /(O) = 

fix) = sec^x. f'iO) = 1 

fix) = 2 sec^x tan x. /"(O) = 

= 2(tana; + tan'x). 

fix) = 2(sec2a; + 3 tan^x sec^a;) /'"(O) = 2 

= 2(1 + 4 tan^x + 3 tan%) 

fix) = 16 tan X sec^x + 24 tan^x sec^x /"'(O) = 

= 16 tan X + 40 tan^x + 24 tan^x 

fix) = 16 sec^x + 120 tan^x sec^x + 120 tan^x sec^x fiO) = 16 

On substituting in (3) we obtain 

X' 2x« 
tan x = x + -^ + Yk''t~'' '• 

The next two terms are ^W ^^ and ^H^ x". 

Exercises 

1. Expand cos x in a power series in x. 

2. Expand cos x in a power series in (x — a). 

3. Expand cos (a+ ^0 in a power series in h. 

4. Expand sin (a + A) in a power series in h. 

6. Express the remainder after three terms in each of the series 
of Exercises 1, 2, 3, 4. 

6. Expand e" in a power series in x. 

7. Expand e"'^'^ in a power series in h. 

8. Expand e" in a power series in {x — a). 

9. Expand log (1 + x) in a power series in x. 

10. Expand log (1 — x) in a power series in x. 

11. Expand tan'^x in a power series in x. 

12. By the use of the series already found, compute: 
(a) ^e to 5 decimal places. 

(6) \^e to 6 decimal places, 

(c) sin 3° to 6 decimal places. 

(d) cosine of 1 radian to 4 decimal places. 

13. By the use of the result of Exercise 3, find cos 33° correct to 4 
decimal places. 

14. By the use of the result of Exercise 4, find sin 32° correct to 4 
decimal places. 

164. Second Proof for Taylor's and Maclaurin's Series. These 



314 . CALCULUS [§164 

series can be obtained very simply in another way if we make 
certain assumptions and do not attempt to justify them. 

Assume that j{x) can be represented by an infinite power series 
in (x — a): 

f{x) = aa-\-ai{x—a)-\-a2{x—aY+' • • +a„(a;— a)"+ • • •, (1) 

where Oo, Oi, 02, ' ' ', On, ' ' ' are coefficients which are to be 
determined. Assume further that the result of differentiating the 
second member term by term any given number of times, is equal 
to the corresponding derivative of the first member. Then, 

f'{x) = ai + 2a2(x — a) -\- 803(0; — a)^ 

+ • • • + nan(x — a)"~^ + • ■ 
fix) = 202 + Ga^ix - a) 

+ ■ ' ■+ n{n - l)a„(a; - a)--^ + ■ ■ ■ (2) 
f"'ix) = 603 + • . • + n(n - l)(n - 2)a„(x - a)"-^ +• • 



/(") (x) = |wo„ + . . . 
Put re = a in (1) and (2). 



whence 



/(a) = ao, 
/'(a) = ai, 
r(a) = 2a2, 
/'"(a) = I3a3, 

/"Ha) = Inan. 

ao = /(a), 
ai = f'(a), 

fia) 
a2 = ^v 

at = —f^ — > 



/("^ («) 

Ctn = , > 



§165] TAYLOR'S AND MACLAURIN'S THEOREMS 315 

Substituting in (1) we obtain 

fix) = f{a) + /'(a)(x -a)+ -^(x - a)' 

By setting a = 0, and x = a + h,we get (3) and (2), respectively, 
of the preceding section. 

165. Tests for the Convergence of Series. Several tests will 
now be given for determining whether or not a series is convergent. 
They will be given without proof, though in most cases the proof 
is not difficult. 

If a series Ui -\- uz -{• • • • + Wn + • • -is convergent, 

lina Un = 0. 
n— 00 

The converse of this statement is not true. Thus the series 

1 + i + i + i + i + • • • + ^ + • • • (1) 



is divergent, although 



^^"^ - = 0. 

n=oo^ 



That this series is divergent can easily be seen as follows: 

3 T^ 4 -^ 2 

"5 + C + Y + "8 > 2 

« ~r 1 0' + 1 f -P "1 2" "T 1 3' + "1 4" "T I 5 "1" 1 6" ^ 2 

The terms of the series can then be grouped into infinitely many 
groups such that the sum of the terms in each group is greater 
than 5. But the series 

is divergent. Much more then is the series (1) divergent. 

Test 1. If ^^"^ Un is not zero the series is divergent. This test 

n= <x> 

is easy to apply and if it shows the series to be divergent, no 
further investigation is necessary. 



316 CALCULUS f§166 

Test 2. Alternating series. A series of decreasing terms whose 
signs are alternately plus and minus and for which 

n— 00 

is convergent. 
Thus the series 

l-l + l-\ + l-ls+ • • ' (2) 

is convergent. 

The reason for the convergence of such an alternating series 
can be seen as follows. Denote by Sn the sum of the first n 
terms and suppose the (n + 1)**^ term positive. (See Fig. 12L) 
Then, since the terms are constantly decreasing, 

Sn + l > Sn', Sn+2 <^ Sn + l', Sn+2 > Sn- 



'Sfl+l ' 



'\^ 



Sn+2 -*{ 

Fig. 121. 

It is clear that as n increases Sn oscillates back and forth but 
always within narrower and narrower limits, owing to the fact 
that the terms are constantly decreasing. As n becomes infinite 
the amount of this oscillation approaches zero since 

71= 00 

Sn therefore approaches a limit. 

Test 3. Comparison Test. If the terms of a series are in numer- 
ical value less than or equal to the corresponding terms of a known 
convergent series of positive terms, the series is convergent. If the 
terms of a series of positive terms are greater than or equal to the 
corresponding terms of a divergent series of positive terms, the series 
is divergent. 

A useful series for comparison is the geometrical series 

a -\- ar -^ ar^ + ar'' + ■ • • + ar" + * * • , (3) 

which is convergent if |r| < 1 and divergent if |r| ^ L See also the 
series (a) of Illustration 2 of this section. 



§165] TAYLOR'S AND MACLAURIN'S THEOREMS 317 

Test 4. The Ratio Test. By comparison with the geometrical 
series it can be shown that the series 



is convergent if 
divergent if 
If 



lim 


Un+l 


n=a> 


Un 


lim 


Un+l 


n=co 


Un 


lim 


Un+l 


«=« 


Un 



<1, 



> 1. 



= 1 



the test fails. In this case other tests must be appHed. 

There are a great many tests for the convergence of series but 
only a few can be given here. It should be added that there is no 
test that can be applied to all cases. 

Illustration 1. Test the series 



1 - t+l-i + 



+ 



for convergence. 
Since 



lii 



Un 



is not zero, the series is divergent (Test 1). It is to be noted that 
the terms of the series are alternately positive and negative and 
that they decrease, but they decrease to the limiting value 1 
instead of 0. Hence test 2 does not apply. 
Illustration 2. Test the series 



1 + -+-+-" + 

^ ~ 9< ' *?« ~ /!< ' 



(a) 



for convergence. This series is useful in testing the convergence of 
series by comparison. 

If f = 1 we have seen that this series is divergent. (See (1).) 
If f < 1 each term of (a) is greater than the corresponding term 
of (1) and hence (o) is divergent. If f > 1 we can compare (a) 
with 

l + l + l + l, + i + ll + l.+ "• (« 



318 



CALCULUS 



r§165 



Each term of (a) is less than or equal to the corresponding term 
of (6). But (6) is convergent since it can be written 



^ + ^(^)+M4^)+M^)-^ 



1+-+-+^+-+ 



which is a geometric series whose ratio, ^~, is less than 1. Hence 

(a) is convergent when t > 1. Summing up: 
(a) is divergent if t ^ 1. 
(a) is convergent U t > 1. 

Illustration 3. Test the series 



1 



1 







i + r2 + r3+i4 + - • • 


for convergence. 


Apply test 4. 


1 


Mn = P- 


jn 


Hence 




1 


lim 


Un+1 


lim 1^ + 1 lim 1 


M=oo 


Un 


1 "=- n + 1 

In 



= 0. 



The series is therefore convergent. 

Illiistration 4. For what values of x, if any, is the series 



^~ 13"^ 15 17 + 



convergent? 



Then 



2n -1 



lim 

n= 00 



Un+l 



Un 





a;2„+i 


im 


|2n + l 


= QO 


a;2„-i 




|2n-l 



lim 

n = oo 



|2n (2n + 1) 



= 0, 



§165] TAYLOR'S AND MACLAURIN'S THEOREMS 319 

for all finite values of x. Hence the series is convergent for all 
finite values of x, positive or negative. 
Illustration 5. For what values of x is 



X'' 

'^3 



X* ,x. 



convergent? 







\Un\ = 


71 












_ lim 


^n + l 


= 


lim 

n= 00 


X 






Un + l 


n+ 1 


n 


Un 


n 


+ 1 






n 













lim 

n=oo 



The series is therefore convergent if |x| < 1. Furthermore it is 
convergent if x = 1 (Test 2), and divergent if x = — 1. (See 
series (1).) 

As has been stated there is no one test of convergence which 
can be applied with certainty of success to any given series. The 
tests which can be most frequently applied have been given. It 
is suggested that the following procedure be observed in general. 

1. See if ^^^Un = 0. 

n= 00 

2. If so, is test 2 applicable? 

3. Tf not, try the fatio test, test 4. This will fail if 



lim 

n= 00 



M„ 



1. 



4. In this case, and in cases where the other tests fail or are 
difficult, try the comparison test. 



Exercises 

Test the following scries for convergence. 

\. \ -\+\ -I + i\ -■ • •■ 

2. i - I + -f - I + A - • • •• 

3. 4 + i + ^ + ^ + A + • • •. 
|3 |4 |5 

10* ^ 

^ A A 

13 + 14+ I 5 + ' ' *• 



5. 



10 "*" 10^ "^ 10' 



1 

12 + 



320 CALCULUS [§166 



^- ^ + 2V2 + 3V3 +4V4 + " ' ■• 
' 32 ^ 52 72 ^9=* 

^' 1-2 ^ 3-4 ^ 5-6 ^ 
For what values of x are the following series convergent? 

10. The Maclaurin's series for e"? Exercise 6, §163. 

11. The Maclaurin's series for cos x? Exercise 1, §163. 

12. The Maclaurin's series for sin x? Illustration 1, §163. 

13. The Maclaurin's series for log (1 — x)? Exercise 10, §163. 

14. The Maclaurin's series for tan~^x? Exercise 11, §163. 

166. Computation of Logarithms. The series of Exercise 9, 
§163, for log (1 + x) is convergent only when —1 <x^ + 1, and 
that for log (1 — x), Exercise 10, §163, only when — 1 ^x< + 1. 
It would appear then impossible to find the logarithm of a number 
greater than 2 by these formulas. By a very simple device it is, 
however, possible to obtain formulas for finding the logarithm 
of any number. 

From the series of Exercises 9 and 10, §163 it follows that 

1 -f J 
log:j = log (1 + x) - log (1 - x) 



X+3-+g-+---], (1) 



where Ixl < 1. Let x = ^ — i~T' Then 
' ' 2z + 1 

1 -hx _ z+ 1 

1 — X z 

z 
where > 0, or 

log(^+l)=log3+2[2^+3(22 + l)3+5(22Vl)^+' '^^^ 

By letting z = 1, log 2 can be computed by this formula. The 
series is much more rapidly convergent than that for log (1 -\- x), 
X = 1. In fact, 100 terms of the latter series must be taken to 



and 

^°^ ~T~ ^ ^L2M^ "^ 3{2z + iy "^ 5(27+Tr^ + • • -J . (2) 



§168] TAYLOR'S AND MACLAURIN'S THEOREMS 321 

obtain log 2 correct to two decimal places, while four terms of the 
new series (3) will give log 2 correct to four decimal places. After 
log 2 has been found, log 3 can be found by setting z = 2. The 
logarithm of 4 is found by taking twice log 2; log 5 by setting 
2 = 4; log 6 by adding log 3 and log 2, and so on. 

Exercise 
Compute log 5 correct to four decimal places, given that log 4 
= 1.38629. Here, as always in the Calculus, the base is understood 
to be e. 

167. Computation of tt. By letting a; = 1 in the series for 
tan~'x. Exercise 11, §163, the following equation is obtained from 
which IT can be computed: 

I = tan-i l = l-i + i-|+- • •• 

This series converges very slowly. To obtain a more rapidly con- 
verging series make use of the relation 

tan~^ 1 = tan"^ ^ + tan~^ |. 
Then 

4 ' (3) (23) + (5) (2^) (7) (2^)^ 

^' (3)(33) ^ (5)(3^) (7)(3^)^ 

168. Relation between the Exponential and Circular Functions. 
If it be admitted that the Maclaurin's series expansion 

e' = l+2 + |2 + ^ + - • •, (1) 

which was proved for real values of z, is also true when z is imagin- 
ary, we obtain, on setting z = ix, 

6.^ = l-hta;+^-h-|3--|-^-h-|5- + ^ + -|7-+- ■ • 

^ , , x^ ix^ , X* , ix^ x^ ix'' , 

= l+tx-j2- || + |4+^-j6--|7 + - • • (2) 

On separating real and imaginary parts this becomes 

e"-l-^ + |4-|g-|---- 



+ 1(2 



13 ' 15 17 
21 



+ %-%+■■). (3) 



322 



CALCULUS 



[§168 



Since (Exercise 1 and Illustration 1, §163), 



and 



cosx = l-j2- + |^-ig + 



x^ , x^ x* . 
sinx=x-j3+^-i^ + 



it follows that 

e« = cos X + i sin X. 

On changing the sign of x it results that 

e-»^ = cos X — 1 sin X. 

Solving equations (4) and (5) for cos x and sin x, 



and 



cosx = 



sinx = 



2 

e** — e- 



(4) 

(5) 

(6) 
(7) 



These interesting relations between the circular and exponential 
functions are of very great importance. 





Fig. 122. 



Fig. 123. 



If Q represent the vectorial angle in the complex number plane, 
then it is clear from Fig. 122 that e*' represents a point on the 
unit circle (circle of radius 1 about the origin as center) in this 
plane. Further, any complex number a + hi can be put in the 
form pe". For (Fig. 123) 

a -^-hi = p (cos + 1 sin Q) = pe*", 

where p = y/a^ + h"^. 



5. e*"-. 


7. 


e^*'. 


6. e-'"". 


8. 


3tV 

5e 4 . 



§169] TAYLOR'S AND MACLAURIN'S THEOREMS 323 

Exercises 

Represent by a point in the complex plane : 

IX IT 

1. 3e^ 3. e^. 

2. 2e" 3 , 4. e^. 
9. Express the numbers of Exercises 1-8 in the form a + hi. 

169. DeMoivre's Theorem. The interesting and important 
theorem, known as DeMoivre's Theorem, 

(cos + i sin 0)" = cos n0 + i sin n^ (1) 

can be easily established by the use of the relation (4) of §168. 

For, 

(cos d + i sin 0)» = {e'^Y = e*"^ = cos nd + i sin nd. 

Exercises 

Find, by the use of (1), 

1. The cube of 1 + i. 1 + iy/^ 

4. Ihe cube of 



2. The square of ^ • - n^i i -. — 1 — iy/Z 

2 5. The cube of ^ -• 

3. The cube of ^ 6. The cube of ^ •• 

In (1), 11 may be a fraction as well as an integer. It will then 

indicate a root instead of a power. In this case we do not have 

simply one root: 

— fl ft 

(cos d -\- i sin 0)'" = cos [- i sin— > 

mm 

(n having been placed equal to — , where m is an integer) but 
m — 1 additional roots. This follows from the fact that 

giO _ gi(0 + 2p») (2) 

where p = 0, 1, 2, 3, 4, • • ■ ,m,m -\- \, • • • . Hence we can write 

(cos d + I sin ^)™ = [e^"]'" = [e'^" + ^p^^^^ 
or 

1^ t(0 + 2pw) 

(cos ^ + t sin d)"" = e »» , (p = 0, 1, 2, • • •) ^^' 



324 CALCULUS [§170 

It would appear at first sight as if there were infinitely many 
roots corresponding to the infinitely many values of p. But a 
little consideration shows that when p y m, the roots already 
found by letting p take the values 0, 1, 2, • • • , w — 1, repeat 
themselves, since e^" = 1. There are then exactly m mM^ roots 
of e*^ = cos + I sin d, 

e- —^ = cos — ' — — + t sm — ^ — —, (4) 

where p = 0,1,2, • • • , w — 1. 

Illustration. Find the three cube roots of — 1. 

(-l)i = (e-)i 

^ ^gtu + 2px)j} (p = 0,1,2) 

t (ir + 2px) 

= e 3 (p = 0, 1, 2) 

tV 5jt 

= e^ , e*', and c ^ . 

Exercises 

ie 

1. Show that the three cube roots of a + 6t = pe'* are : -y^ « ^i 
«(g + 2t) i(g -f 4t) 

■^^ e 3 J and -^^ c 3 . How would these roots be deter- 
mined graphically? 

2. Find the two square roots of 1 + i. 

3. Find graphically the two square roots of l. 

4. Find graphically the three cube roots of 1. 

170. Indeterminate Forms. It has already been shown that 

^ has no meaning. See §25. Thus 

x-2 
has no meaning at a; = 2. Its value at x = 2 is defined as 

lim ^^ ~ '^ = 4 

x=2 X — 2 

Similarly 

sin a 
tan a 
has no meaning at a = 0. Its value at a = is defined as 
lim sin a ^ ^ 
°=o tana 



§170] TAYLOR'S AND MACLAURIN'S THEOREMS 325 



In general, if </)(a) = and /(a) = 0, the value of the function 
lim ^xji 

x = a f(^x) ' 



-77-v Sii X = a IS denned as 



The calculation of this limit is simplified in many cases by the 
application of the law of the mean. See §160. 

Thus, let it be given that 4>{a) =0 and /(a) = and let 4)(x) 
and f(x) satisfy the conditions imposed in the statement of the 
law of the mean. Then 

lim 0(a;) ^ Hm </>(») + (x - a) 4>' [a + 0\{x - a)] 
x^a /(^) x^a ji^a) + {x-a) f [a + d^ix - a)] 

__ lim 4>'[a + ei(x- a)] 0^(a) ^ .^i . . 
- z^a f'[a + e,{x-a)] /'(a) ' "^ ^6,^ ^^ 

If (f>'(a) and /'(a) are also zero, we make use of the extended law 
of the mean. Thus 

0(a) + (x - a) 0'(a) + ^^-^V"[a + ^1(0;- a)l 
lim 4>{x) ^ lim |i 

•'^ ^ /(«) + (x- a) f'(a) + ^-^—^ria + d^ix - a)] 

_ lim </> '[a + gi (x - g)] <i>"(a) 
x^a j"[a + d,{x - a)\ f{a) ' 

The process is to be continued further if f"(a) and <^"(a) are 
both zero. 

Illustration 1. 

lim e' -1 ^ 

x=o X 

Illustration 2. 

lim e^ — e~' — 2x 



a:=&0 



X — sm a: 



Hm e- 

z=0 1 


= 1. 




lim € 


. + e-x_ 


2 


' x=0 


1 — cos X 




lim e^ 


. _ g-x 




x=0 


sin X 




lim e' 


+ e-* 




1=0 


p.na X 





2. 



326 CALCULUS [§170 

The Form — . The same process is employed in evaluating 

00 

the indeterminate form — -. The proof is omitted. 

Illustration 3. 

lim ^ _ lim 2x _ lim 2^ _« 

X— 00 nx X— oo gx X= 00 ox ' 

The Form <» . The indeterminate form «> can be thrown 

00 

into either of the forms - or — ■. Thus 

00 

lim 3. „„. y _ lim _^ lim 1 _ . 

=^=0 -^ ^"^ ^ ^=0 tan X ^^0 sec2 x~ 

Other indeterminate forms are : °o — «> , 1 °°, 0", °°°. 
Thus, if <p(x) and /(x) become infinite for x == a, (/>(a) — /(a) is 
defined as 

^^^JHx) - fix)]. 

This expression can be written 

1 1 

lim r , / V ,/ ^^ _ lim fix) <t>{x) 

W)f(x) 
an indeterminate form of the type ^. 

If (l>{x) becomes infinite and/(x) becomes 1 for x = a, [/(a)]*^"^ 
is defined as 

This limit can be calculated as follows. Let y = [/(x)]*^^'. 
Then 

logy = 4){x) log/(x) = ^ > 

W) 
an indeterminate form of the type -. If 

lim log /(a;) 

W) 



§170] TAYLOR'S AND MACLAURIN'S THEOREMS 327 
is found to be c, then 



lim „ = g« 

x = a 



The two remaining forms are evaluated in a manner similar to 
the last. 

Many indeterminate forms can be evaluated directly by simple 
algebraic transformations. 



Exercises 



Evaluate the following: 

lim loga^ 13 lim _^. 

J- x=l X - i • 2=°' log a; 



lim 1 -cosg 14. lim ^ 

lim c» 



9=0 cosflsin^fl 



« lim a; COS a; — sin x 16. ^^ „ ^ 



X = X 

4. 



lim tan x - sin x 16. ^™ e« tan -• 

z=0 a; — sin X 

lim a:" - 1 



^- a;=l X - 1 ^„ lim 



lim sin 33; 
• 1=0 sin 2a; 



17 lim r_^ L_" 

^'' X=l Lx2 - 1 X-l_ 

i™ [(|-9)ta„; 

2 

- i=u smzx r r 1 11 

^ lim tan 3a; 19. ^^^ |^j^ - a._ij' 

g Um ^^ 20.1;- (cos xfot^. 
x=o sin X — a; 1 

lim x^ - X - 6 21. iTod-^)^- 
^- x=3 x2 — 9 i:,„ 

,„,ta 311+5 22.'-(-x).»«. 

• I* . 4i! + 1 jj_ hm (3j„ ^)t.n . 
lim log X 

12. li™ '^- „. lim tan fl. 



CHAPTER XVIII 

TOTAL DERIVATIVE. EXACT DIFFERENTIAL 

171. The Total Derivative. Let z = f{x, y) and let x and y be 
functions of a third variable t, the time for example. We seek an 

dz 
expression for -rr, the derivative of z with respect to t, in terms of 

dx J dy 

As an illustration of what is meant, let z denote the area of a 
rectangle whose sides x and y are functions of t, and at a given 
instant let each side be changing at a certain rate. The rate at 
which the area is changing is sought. 

Returning to the general problem let t take on an increment At. 
Then x takes on the increment Ax and y the increment Ay, and 
consequently z the increment Az. We then have 

2 = fix, y) (1) 

2 -f Az = /(x + Ax, 2/ + Ay) 

Az=J{z + Ax,y + Ay) - /(x, y) (2) 

Az = /(x+Ax, y+Ay)-J{x, y+Ay)+f{x, y+Ay)-f(x, y) (3) 
. Az ^ fix -\-Ax,y + Ay) - /(x, y + Ay) Ax 
At Ax At 

fix, y + Ay) - fix, y) Ay 
"•" Ay At ^ ' 

Taking the limits of both sides of (4) as At approaches zero, we 
have 

dz _ dfjx, y) dx dfjx, y ) dy , . 

dt dx dt "^ dy dt' ^ ' 

since Ax and Ay each approach zero as At approaches zero. 
Equation (5) can be written in the form 

dz _ 3z dx 3z dy , . 

dt ~ ax dt "•" ay dt * ^^ 

328 



§171] TOTAL DERIVATIVE. EXACT DIFFERENTIAL 329 

This states that the rate of change of z with respect to t is equal 
to the rate of change of z with respect to x, times the rate of change 
of X with respect to f, plus the rate of change of z with respect to 
y, times the rate of change of y with respect to i. 
\il = Xy (6) becomes 

dz 

dx 



dz dz dy 
dx dy dx 



This formula applies when z = f{x, y) and ?/ is a function of x, 
e.g., y = <i>{x). 

Multiplying (6) by dt we obtain 



dz 



^r dx + ^- dy. 
dx dy 



(7) 



This defines dz, which is called the total differential of z. 



z 


E 


I 


S 

c 




F 
R 
A 


f 


K 


% 


7 






Y 






dy 








r 




/ 





Fig. 124. 

We shall now give a geometrical interpretation of dz. Let 
P, Fig. 124, be the point {x, y, z) on the surface z = fix, y). 
Let 

PC = dy 
and 

PA = dx. 

Then Q is the point {x + dx, y + dy, z + Az). Let PDEF be 
the plane tangent to the surface at the point P. Then PF is 
tangent to the arc PR, and PD is tangent to the arc PS. 



330 CALCULUS [§171 

From F draw FK parallel to AB meeting BE in K. 

BE = BK-\- KE 

BK = AF = ^ dx. 
ox 

Since FK = PC and PD = FE, triangle KFE is equal to the 
triangle CPD, and 

KE ^ CD = ^ dy. 
dy ^ 

Therefore 

^^ = dx^ + 3y^y' 

Hence BE = dz. Consequently dz may be interpreted as the 
increment measured to the tangent plane to 2 = f{x, y) at the point 
P (x, y, z) when x and y are given the increments dx and dy 
respectively. 

Illustration 1. \i z = xy, the area of a rectangle of sides x 
and y, we obtain by using (7), 

dz = y dx -\- X dy. 

The first term on the right-hand side represents the area of the 
strip BEFC, Fig. 125. The second term the area of DCGH. The 

difference between Az and dz is the 
^ area of the rectangle CFLG, which 



becomes relatively smaller, the 
smaller dx and dy become. 

The above expression could have 
been obtained by the formula for the 
B dx E differential of the product of two 
Fig. 125. variables. 

Illustration 2. The base of a 
rectangular piece of brass is 15 feet and its altitude is 10 feet. 
If the base is increasing in length at the rate of 0.03 foot per hour 
and the altitude at the rate of 0.02 foot per hour, at what rate is 
the area changing? 

Let X denote the base, y the altitude, and z the area. 
Then 

z = xy 



§171] TOTAL DERIVATIVE. EXACT DIFFERENTIAL 331 

and dz _ dx dy 

dt~'^ dt '^^ dt 

= (10) (0.03) + (15) (0.02). 

X 



Illustration 3. z = -• 

y 






^ _ 1, 

dx y 


dz 

dy 


X 

~ y'- 


and, by (7), 







dz = — dx 5 dy, 

y y^ ^' 

or 

, y dx — x dy 

dz = 5 > 

yi 

a result which could have been obtained by differentiating the 

X 

quotient - by the usual rule. 

Exercises 

Find by formula (7) the total differential of each of the following 
functions : 



1. z = x^y. 




-=^.- 




7. 2 = x'e". 


2. 2 = XJ/^ 




6. 2 = z log y. 




S. z = e* sin y. 


a;2 
3. 2 = — 

y 




6. z = e* cos y. 




9. 2 = e"" cos nx. 


Find ^f if: 
dt 










10. z = X* cos 


y- 


11. 2 


= e* 


sin y. 



12. The radius of the base of a right circular cylinder is 8 inches and 
its altitude is 25 inches. If the radius of the base is increasing at 
the rate of 0.2 inch per hour and its altitude at the rate of 0.6 inch 
per hour, at what rate is the volume increasing? 

13. Given the formula connecting the pressure, volume, and tem- 
perature of a perfect gas, pv = Rt, R being a constant. 11 t = 523°, 
p = 1500 pounds per square foot, and v = 21.2 cubic feet, find the 
approximate change in p when I changes to 525° and v to 21.4 cubic 
feet. 

14. If with the data of Exercise 13, the temperature is changing at 



332 CALCULUS [§172 

the rate of 1° per second, while the volume is changing at the rate of 
0.4 cubic foot per second, at what rate is the pressure changing? 

15. The edges of a rectangular parallelopiped are 6, 8, and 10 feet. 
They are increasing at the rate of 0.02 foot per second, 0.03 foot per 
second and 0.04 foot per second, respectively. At what rate is the 
volume increasing? 

172. Exact Differential. An expression of the form 

M dx -{- N dy, 

where M and N are functions of x and y, may or may not be the 
differential of some function of x and y. If it is, it is called an 
exact differential. Thus 

sin y dx -\- X cos y dy (1) 

is an exact differential, for it is the differential of z = x sin y. 

The coefficient of dx is ^ = sin y, and that of dy is \- = x cos y. 

x^ sin y dx -\- x cos y dy (2) 

is not an exact differential. It is fairly evident from (1) that we 

dz 
cannot find a function z = f{x, y) such that v- = x^ sin y and 

dz 

-^ — X cos y. 

dy ^ 

We seek to find a test for determining whether or not an ex- 
pression of the form 

Mdx + N dy (3) 

is an exact differential. If (3) is the exact differential of a func- 
tion z, we must have, 

| = M (4) 

and 

1 = ^. (5) 

since 



§172] TOTAL DERIVATIVE. EXACT DIFFERENTIAL 333 

Differentiate (4) with respect to y and (5) with respect to x 
and obtain 

AU AM 

(7) 
(8) 





dh dM 


and 


dy dx ~~ dy 
d^z dN 


Since, in general, 


dxdy " dx 
d^z dh 




dydx ~ dxdy 



it follows that if (3) is an exact differential, we must have 

dy dx 
The condition (9) must be satisfied if (3) is an exact differential. 
It does not follow, however, without further proof, that (3) is 
an exact differential if (9) is satisfied. It can, however, be shown 
that this is the case. The proof will be omitted. (3) cannot be 
an exact differential unless (9) is satisfied and is an exact differ- 
ential if (9) is satisfied. 

When an expression of the form (3) is given, the first step is to 
determine whether or not it is an exact differential by applying 
the test (9). If it is an exact differential, the next step is to 
find the function z of which it is the differential. This step will 
be illustrated by integrating several differentials for which the 
functions from which they were obtained by differentiation are 
known. 

Illustration 1. If z = a;^ + 2xhj -\- y^ -\- C, 

^' = dic'^'' + dl^y 

= {3x^ + 4xy)dx + (2x^ + 2y)dy. 
If then we are given the exact differential 

dz = (3a;2 + ^xy)dx + {2x^ + 2y)dy 
and arc required to find the function of which it is the differential, 
we note first that 



dz 
Then 



d^x = ^^' + ^^2/. 



2 = a;3 _|_ 2xhj + a Junction of y alone. 



334 CALCULUS [§172 

And this function of ?/ is to be so determined that 
dz 

Clearly the term 2x^ is obtained by taking the derivative with 
respect to t/ of 2x^2/, a term already found, and consequently it 
is not to be added. 2y is the derivative of y^. y^ is then the 
function of y which is to be added to the terms already found. 
Further an arbitrary constant is to be added since its differential 
will be zero. Then 

z = x' + 2xh/ + y^ + C 

is the function whose differential was given. If, as is usually the 
case, it had been given that 

(3x* + ^xy)dx + (2x2 + 2y)dy = (10) 

it would have been required to find a function of x and y such that 
its differential would be zero. Now the first member is, as we 
have seen, the differential of 

z = x^ + 2xh) + y\ 
But, if (fe = 0, 2 = C. Then 

x» + 2xhi + y^ = C 

is the relation between x and y which satisfies the given equation. 
Illustration 2. If 

z = e* cos y -{- x^ -\- sin y + y^, 

dz = (e* cos y -\- 2x) (fx + ( — e* sin y + cos y + ^y^) dy 

Now let it be given that 

(e* cos y + 2x) dx -\- (— e' sin y -\- cos y + Sy^) dy = 0. (11) 

From its derivation we know that the left-hand member is an 
exact differential, dz. Let us proceed to find z as if it were 
unknown. 



dz 
Then 



-— = e" cos y + 2x. 



z = e* cos y + x^ + a function of y alone. (12) 



§173] TOTAL DERIVATIVE. EXACT DIFFERENTIAL 335 

The function of y is to be so determined that 

dz 

^ = - e' sm y + cos rj -{■ dy'^. (13) 

The first term is evidently obtained by differentiating e' cos y, a 
term already found in (12). The remaining two terms in (13) 
are obtained by differentiating sin y + y^. These are to be added 
to the terms already found in (12). 
Then 

z = e' cos y -{- x^ -\- sin y + y^. 
But, since dz = 0, z = C. Hence 

e" cos y -{- x^ -{- sin y -\- y^ = C 

is a solution of (2). 

Illustration 3, Integrate if possible the equation 

{e' y+sin y-\-2x) da;+(e'+x cos y-\-e''-\-2y—sin y) dy = 0. (14) 

We have first to determine whether or not the first member is 
an exact differential. Apply the test (9), 

dM 

~ = e' + cos y. 

dN 

—— = e* + cos y. 

Hence (9) is satisfied and the first member of (14) is an exact 
differential. On integrating the coefficient of dx with respect to 
X we obtain 

c^y -{- xsiny -{- x"^. 

To this we have to add 

gV ^ y2 _|_ pQg y^ 

the terms which arise from the integration of the coefficient of dy 
and which contain y alone. (The other terms in the coefficient of 
dy arise from the differentiation of terms already found by integrat- 
ing the coefficient of dx.) Then the solution of (14) is 

e^y + xsiny + x^ + e" + y^ + cosy = C. 

173. Exact Differential Equations. Equations involving differ- 
entials or derivatives are called differential equations. Those of 
the type 

Mdx + N dy = (1) 



336 CALCULUS [§174 

where the first member is an exact differential, are called exact 
differential equations. 

The equations (10), (11), and (14) of Illustrations 1, 2, and 3, 
§172, are exact differential equations. The process of finding 
the relation between y and x, which when differentiated gives a 
certain differential equation, is called the integration of the 
equation. 

The procedure in dealing with an equation of type (1) is to 
determine first whether or not it is exact by applying the test (9), 
§172. If it is, integrate the coefficient of dx with respect to x 
and to this result add those terms which contain y only, which 
are obtained by integrating the coefficient of dy with respect to y. 

Exercises 

Are the following differential equations exact? Integrate those 
which are exact. 

1. Sx'^y^dx +2xhjdy = 0. 

2. — cos(-)dx cos (-) dy = 0. 

y \yl y' \y/ 

3. y e'" (1 + X + y) dx + X e'" {I + x + y) dy = 0. 

4. y e'" dx + X e'" dy = 0. 

5. {x^y + 2x) dx - {Sx^y - 5x) dy = 0. 

6. (p+l)dx- [^ + 2y)dy=0. 

7. e-y(2 + -^^ - j^ e^2 + ^^dy=0. 

174. In §155 the envelope of a family of curves was defined, 
and its parametric equations were found to be 

fix, y, c) =0 (1) 

We shall now show that the envelope is tangent to each curve 
of the family of curves (1). 

At a given point {x, y) of the curve determined by giving c 
a particular value in (1), the slope of the tangent is found from 
the equation 

dj.dfdj,^^ 
dx dy dx ' 



§174] TOTAL DERIVATIVE. EXACT DIFFERENTIAL 337 

If the point also lies upon the envelope its coordinates satisfy 

(1) and (2). The equation of the envelope can be regarded as 
given by (1) where c is the function of x and y found by solving 

(2) for c. On differentiating (1) with respect to x, regarding c 

as a function of x and y, the slope, -r-, of the tangent to the en- 
velope is given by 

dx dy dx dc dx ' 

dc _ dc dc dy 
dx dx dy dx 



where 



fif 

But on the envelope ^ = 0. Hence (4) becomes 

^ + ^ ^ = (5) 

dx dy dx ' 

Equations (3) and (5) show that the slope of the tangent line to 
the envelope at the point (x, y) is the same as the slope of the tan- 
gent line at the same point to a curve of the family (1). Hence 
the envelope is tangent to each curve of the family of curves (1). 



22 



CHAPTER XIX 
DIFFERENTIAL EQUATIONS 

175. Differential Equations. An equation containing deriva- 
tives or differentials is called a differential equation. If no deriva- 
tive higher than the first appears it is called a differential equation 
of the first order. If the equation contains the second, but no 
higher derivative, the equation is said to be of the second order. 
And so on. Numerous differential equations have already 
occurred in this course. We shall now consider the solution of 
differential equations somewhat systematically. 

176. General Solution. Particular Integral. Let 

fix, y,c)=0 (1) 

be any equation between x, y, and the constant c. If (1) is 
differentiated with respect to x there results the equation 

Fix, y, y', c) = 0. (2) 

Between (1) and (2) the constant c can be eliminated giving the 
differential equation of the first order 

<l>{x, y, y') = 0. (3) 

Equation (3) follows for any value of the constant c. 

Let 

/(x, y, ci, C2) = (4) 

be an equation involving two constants, Ci and Co. By differ- 
entiating (4) we obtain 

Fix, y, y', ci, d) = (5) 

and 

<t>ix, y, y', y", ci, c^) = 0. (6) 

Between equations (4), (5), and (6), Ci and C2 may be eliminated 
giving the differential equation of the second order 

Hx, y, y', y") = 0. (7) 

338 



§178] DIFFERENTIAL EQUATIONS 339 

From the equation (1) containing one arbitrary constant the 
differential equation of the first order (3) is obtained. From the 
equation (4) containing two arbitrary constants the differential 
equation of the second order (7) is obtained. In like manner 
from a relation between x and y containing n arbitrary constants 
a differential equation of the n*^ order is obtained by differentiat- 
ing, and eliminating the constants. 

Equation (1) is a solution of equation (3). It is called the 
general solution and involves one arbitrary constant of integration, 
c. Equation (4) is called the general solution of (7). It involves 
two arbitrary constants, or constants of integration. It can be 
shown that the general solution, or general integral, of a differential 
equation contains a number of arbitrary constants, or constants 
of integration, equal to the order of the differential equation. 

A particular integral is obtained from the general integral by 
giving particular values to the constants of integration. 

177. Exact Differential Equations. This type of differential 
equation was discussed in §173. 

178. Differential Equations; Variables Separable. The vari- 
ables X and y are said to be separable in a differential equation 
which can be put in the form f(x) dx + </>(?/) dy = 0. The first 
member is equal to a function of x alone multiplied by dx plus a 
function of y alone multiplied by dy. 

Illustration 1. 

(1 + y'^)x dx-\- {l-\- x^)y dy = 0. 

On dividing by (1 + ?/^)(l + x^) this equation becomes 

xdx y dy _ 



l + x^ ' 14-2/ 
Integration gives 

ilog (1 + x^) + ilog (1 + 2/2) = C. 

This reduces to 

'1> 



(1 + x^){\ + 2/2) = e2C = c. 



or 



340 CALCULUS I§179 

Illustration 2. 

Vl - y^ dx + \/l - x2 rf?/ = 0. 
Then 

dx dy ^ ^ 

Vl -ic2 Vl - 2/' ' 

and the variables are separated. Integration gives 

sin~i X + sin~i y = C. 

Take the sine of each member, observing that the first member is 
the sum of two angles, and obtain 

a;Vl - 2/' + yVi - x"" = sin C = d. 

Exercises 

Solve the following differential equations: 

1. (1 - x)dy - (1 + y)dx = 0. Ans. (1 + y){l - x) = C. 

2. sin X cos y dx = cos x sin y dy. 

8. (x - Vl + x^) Vl + 2/^ dx = (1 + x^)dy. 

4. ^^ = 5y^x. 

f. dy y" + 4?/ 4- 5 _ 
**• dx "^ x2 + 4x + 5 ~ 

6. (1 + a:)dj/ = y(l — i/)dx. ^ns. ?/ = c(l + x)(l — y). 

7. (1 - x)ydx + (1 + j/)x dy = 0. 

dv 

8. ^ + e'2/ = e'y«. 

9. (x» + j/x*)di/ - (t/2 - x?/2)dx = 0. 

dy , ^ dy 

11. 3e* sin y dx + (1 — e') cos y dy = 0. 

12. (xy + x»2/)dy - (1 + y^)dx = 0. 

Ans. (1 4- x')(l + y') = ex*. 

179. Homogeneous Differential Equations. The differential 
equation 

Mdx-\-N dy = Q (1) 

is said to be homogeneous if M and iV are homogeneous functions 
of X and y of the same degree. 



§179] DIFFERENTIAL EQUATIONS 341 

A function f{x, y) of the variables x and y is said to be homogen- 
eous of degree n if after the substitutions x = Xx', y = \y' have 
been made, 

fix, y) = X-/(x', y'). 
Thus 

ax^ + bxy -\- cy^ 

is liomogeneous of degree 2. For, on making the substitutions 
indicated, it becomes 

\\ax'^ + bx'y' + cy'^). 
The expression 



ax^Vx^ + 1/2 + 6x3 tan-» (-\ 



is homogeneous of degree 3. For, after the substitutions indi- 
cated above, it becomes 

X3 fox'^^/x'^ + y'^ + 6x'3 tan-i Z^;) ] • 

A homogeneous differential equation of the form (1) is solved by 
placing y = vx, and thus obtaining a new differential equation 
in which the variables, v and x, are separable. 

Illustration: 

(x2 + 2/2) dx + 3xy dy = 0. 
Let 

y = vx. 
Then 

dy = V dx -\r X dv, 
and 

x2(l + v^) dx + Svx%v dx -\-x dv) = 0. 
x2(I + 4«;2) dx + 3^3 dv = 0. 



Separating the variables 



dx Svdv _ 
T "•" 1 + 4i;2 " "• 
log[x(l +4i;2)i] = C, 

x(l 4-4i'2)i = Ci. 



342 CALCULUS [§180 

y 

On substituting t; = - we obtain as the solution of the given 
equation 

or 

x\x^ + 4?/2)3 = Ci. 

Exercises 

Solve the following differential equations 

^ _ ^„ #. 
dx ~ ^y dx 

2. x'^y dx - (x' + 2/3) di/ = 0. 

3. (8y + 10a;) dx + (5y + 7x) dy = 0. 

4. (2\/x?/ — x) dy + 2/ dx = 0. 

?/ ^2/ 2/ 

6. X cos — -J- = y cos X. 

X dx X 

dy 



7. x ^ - y = Vx'' - y\ 

8. {y - x) dy + ydx = 0. 

180. Linear Differential Equations of the First Order. The 

equation 

where P and Q are functions of x only, is called a linear differential 
equation. It is of the first degree in y and its derivative. Multi- 
ply the equation by 

' Pdx 



and obtain 



./' 



iPdxrdy , „ 1 {pdx ^ 



(2) 



The left-hand member is the derivative of 

fpdx 
eJ y, 

as may be confirmed by differentiating this product. The inte- 
gration of (2) gives 



}180] 



DIFFERENTIAL EQUATIONS 343 



e* 


f'-"v = JQeJ'""dx + C. 


Illustration 1. 






t+^y-"'- 


Here P = x and Q 


= x\ Then 




I Pdx \ xdx T 


X- 

Multiply both members of (3) by c 2 . 






Integration gives 


a;2 /. a;2 




a;2 a;2 

= e^x^ - 2e2 + C. 


Hence 


a;2 




J/ =x2-2 + Ce~"^. 


Illustration 2. 





I Pdx (^ 

oJ = P. '' = 



Multiply both members of (4) by x. 

Integration gives 

xy 
or 



= ^ + x^* + 2x2 + C, 



(3) 



^ + l^=:.2 + 3^+4. (4) 

dx x*^ 



x' C* 

y = ^ + x24-2x + -- 

This illustration is inserted to call attention to the well-known 
simple relation e'"«^ = x, which there will be frequent occasion 



344 CALCULUS [§181 

CO use in solving equations of this type. It should be recalled 
that e''^°«^ = e'°«(^"> = x". Thus 











e 


— loga; _ t 
X 

Exercises 


1. 


dy 
dx 


+ 2xy = e 


■^\ 






2. 


dy 
dx 


+ y cos X = 


sin 


2x. 





. dy , 

3. cos'^x-T — r y = tan x. 

4. (x« + l)^ + 2xy = 4x2. 

'•g + ^ = (- + «•• 

6. x(l - x«) dy + (2x2 _ i)^ ^^ = ^x^ ^^.^ 

„ dy y 
1. -/ -n- = e'x". 
dx X 

8. (1 + x^) dy ^- (xy -^ dx = Q. 

^ dy ,1 -2x 

10. (1 + y"^) dx = (tan-iy - x) dy. 

181. Extended Form of the Linear Differential Equation. An 
equation of the form 

% + Pv = Qr (1) 

is easily reduced to the linear form. For, on dividing (1) by y, 
we obtain 

The first term of the left-hand member of (2) is, apart from a 
constant factor, the derivative of ?/""+', which occurs in the second 
term. If we let z = 2/~"^^ we obtain the linear differential 
equation 

or 

g _ (n - l)Pz = -in- 1)Q. 



§1821 DIFFERENTIAL EQUATIONS 345 

Illustration 1. 



dv 

T — h 2/ cos X = y* sin 2x. 



Dividing by y* 
Let 2 = 7/-3. Then 



.dy,, . „ 

y~* -j~ + y~^ cos X = sin 2x, 



,-4 ^ = _ 1 ^ 



and the equation becomes 
dz 
dx 



^ dz , . „ 

— 5 ;i — \- z cos X = sin 2x, 



or 

3 Zz COS X = — 3 sin 2a;. 

dx 

This equation can be readily solved by §180; y~^ is to be sub- 
stituted for z in the result. 

Exercises 

1- ^- + ~ 2/ = a;V. 4. (1 - x2) ^ - xy = axy^. 



dy 
2. ;^^ + 2/ = xy3. 


^'t+ly = '^'y^- 


3.3,^1-7,3 = 


X + 1. 6. X ^ + , = ,2 log X. 




7 '^^ + 2 , - ^'. 



182. Applications. Let there be an electric circuit, whose 
resistance is R, whose coefficient of self-induction is L, and which 
contains an electromotive force, which at first we shall suppose 
constant and equal to E. It is required to find the current i at 
any time t after the time < = 0, at which the circuit was closed. 
The equation connecting the quantities involved is readily set up. 
The applied E.M.F., E, must overcome the resistance of the circuit 
and its self-induction. The former requires an E.M.F. equal to 
iR, and the latter an E.M.F. proportional to the time rate of 

dz dz 

change of current, viz., j, and equal to L ^. The applied E.M.F., 

E, must equal the sum of these two E.M.F.'s. 



346 CALCULUS [§182 

Hence ^ di , ^. ^ ,,. 

Lj^ + Ri = E, (1) 

The student will show that, if i equals zero when t equals zero, 
the solution of this linear equation is 

Er ^h 

If the battery or other source of E.M.F. is suddenly cut out of 
the circuit, the current falls off in such a way that the differential 
equation 

Lj^ + Ri = (3) 

is satisfied. Show that the law at which the current falls off is 

t = ioe-L^'- '»>, (4) 

if the instant at which the battery is cut out is the time 1 = 1^ 
and if the current at this instant is i = Iq. 

If the E.M.F. is variable, the relation between the quantities 
involved in the circuit is still governed by (1), 

Lj^ + Ri^E, (1) 

in which E is now variable. Suppose E = Eo sin cof. This sup- 
poses that an alternating E.M.F. is acting in the circuit. The 
differential equation to be solved is 

di 
L-r. + Ri = Eo sin cat. (5) 

Show that 

*« ~ L R , 



R^ + co^L 



y- sin cat — 03 cos coHe ^ + C 

ft 
(R sin cat — caL cos cat)e + C 



Eq . , . , X ^ 



^f 



V<R' + w^L^ 



=r-rr- siu (cof — 0)e + C, 



§183] DIFFERENTIAL EQUATIONS 347 

where 

sm <f) = 



Then, 



cos d> = — ,- =. 



R 



Jp — ^ t 

i = ^ ° sin (co< - 0) + Ce ^ 



Since the last term becomes negligible after a short time because 
of the factor 

e ^ > 

it is scarcely necessary to determine C. On droppmg out the 
last term as unimportant except in the immediate vicinity oit =0, 
we have 

The current, therefore, alternates with the same frequency as the 
E.M.F., but lags behind it and differs from it in phase by 4>- It 

is to be noted that the maximum value of the current is not -^5- 

K 
JP 

but . • The quantity -s/R^ + oj^L^ replaces, in alter- 

V /2^ + co^L^ 

nating currents, the resistance R of the ordinary circuit. It is 

called the impedance of the circuit. 

183. Linear Differential Equations of Higher Order with Con- 
stant CoeflScients and Second Member Zero. A typical differen- 
tial equation of this class is the following: 

d"y , d^-^y , d"~hj , , dy , ^ /,s 

where Oq, Oi, • • •, a„ are constants. As the equations of this 
class which occur in the applications are usually of the second order 
we shall confine our discussion in this article to linear differential 
equations of the second order. Consider 



348 CALCULUS [§183 

Let us assume that 

2/ = e-"* (3) 

and find, if possible, the values of m for which (3) is a solution of 
(2). The substitution of (3) in (2) gives 

e"" {aom^ + aim + a^ = 0. (4) 

The first factor cannot vanish. The second, equated to zero, gives 
a quadratic equation in m. Call its roots nii and mt. Then (3) 
is a solution of (2) if m has either of the values Wi or mo, the roots 
of 

OqW^ + aim + a2 = 0. (5) 

The equation (5) in m, obtained from the given differential equa- 

d'^v du 

tion by writing m^ for y-^ and to for -v- is called the auxiliary 

equation. 
Two solutions of (2) are 

y = gmii and y = e'^i*. 
Furthermore, 

y = Cie-"!" 

is a solution of (2). For, after the substitution of this value of 
y in (2), Ci can be taken out as a common factor and the other 
factor vanishes in accordance with (4) or (5). In the same way, 

y = de'^i' 

is a solution of (2). And finally the sum of the two solutions 

y — Cie""!^ + C2e"'2* (6) 

is a solution of (2). This can be seen by substituting in (2) and 
recalling that nii and mo are roots of (5). When mi is not equal 
to m2, (6) is known as the general solution of the differential equa- 
tion (2). It contains two arbitrary constants, the number which 
the general solution of a differential equation of the second order 
must contain. 

The values of these constants are determined in a particular 
problem by two suitable conditions. 

Illustration. 



§184] DIFFERENTIAL EQUATIONS 349 

The auxiliary equation is 

m^ — bm + 6 = 0, 
[m - 2){m - 3) = 0. 

Hence mi = 2, mz = 3. The general solution is then 

y = Cie^' + C26''. 





Exercises 


»-g-l-''-°- 




2.g-4, = 0. 




»-g-4^'»-»- 




*-g + |->^-»- 




'■2+^t-o- 





184. Auxiliary Equation with Equal Roots. The method just 
given fails when the auxiliary equation has equal roots, mi = W2. 
For, equation (6), §183, becomes 

y = de""!' + C2e'"2* 
= (Ci + C2)e-^'. 

But Ci + C2 is an arbitrary constant and the solution contains 
only one arbitrary constant instead of two. When the auxiliary 
equation has equal roots, mi = 7^2, equation (2) can be written 
in the form 

Its general solution is 

This solution can be verified by direct substitution. 
Illustration. 

^_4^+4v = 0. 
dx^ dx^ y "• 

The auxiliary equation is 

m^ — 4m + 4 = 0. 
TMi = m2 = 2. 



350 CALCULUS [§185 

The general solution is 

2/ = (C, + C^x)e^'. 

Exercises 

l-^^ + '^g + o-"- 

185. Auxiliary Equation with Complex Roots. If the auxiliary 
equation has complex roots the general solution can be written 
in a form different from (6), §183. The importance of the result 
will be evident at once when it is observed that it contains the 
harmonic functions sine and cosine. If the coefficients of the 
given differential equation (2), §183 are real, and if mi and niz 
are complex, they must be conjugate imaginary numbers. Let 
THi = a -\- ih. Then mz = a — ih. Then (6) becomes 

y = Cie" + '*' + Cze "' ~ *'" 

Now, by (4) and (5), §167, 

gift* = cos bx + i sin bx 

Q-ibx — COS bx — i sin bx. 
Then 

y = c* [(Ci + C-i) cos bx + i{Ci — Cz) sin 6a;] 

On placing Ci + d = A and i{Ci — Cz) = B, we obtain 
y = go* (^ cos bx -\- B sin bx) 
= e"' C cos (6a; — <f>). 

In the last form the two arbitrary constants of integration are 
C and <f). 
Illustration 1. 

The auxiliary equation is 

m2 + 4m + 13 = 0. 



§186] DIFFERENTIAL EQUATIONS 351 

I Hence 

m = -2 ± Si. 
Then 

y = e-2x (J^ COS Sx -\- B sin 3a;). 

Illustration 2. 

2^ + ^y = o. 



Whence 
Then 



m^ + 4 = 0. 
m = ± 2i = + 2i. 
T/ = A cos 2x + i5 sin 2x. 
Exercises 






d^ 
d^9 

5^ + .^. = o. 



2.^, + 9y = 0. 



dx^+'^ = 0- 



186. Damped Harmonic Motion. The resistance offered by the 
air to the motion of a body through it, is roughly proportional to 
the velocity, if the velocity is a moderate one. In §81, the differ- 
ential equation of the motion of the simple pendulum was derived 
on the assumption that the force of gravity was the only force 
acting upon the bob of the pendulum. If the resistance of the air 
is also taken into account we shall have to add to the second 

member of the equation, I -xr^ = — ^ sin 0, a term, — 2kl -r. pro- 

portional to the velocity I -rr- See equation (1), §81. The differ- 
ential equation of the motion is then 

idW . ^ ^,^dd ... 

l^=-g.me-2kl-^. (1) 

The negative sign is used before the last term because the force 
due to the resistance of the air acts in a direction opposite to that 



352 CALCULUS I §186 

of the motion. The advantage of choosing 2k as the proportion- 
ality factor instead of k will appear later. A; is a positive constant. 
From (1) we obtain 

j;|+2.f + f.„. = 0. (2) 

As in §81 assume that 6 is small and replace sin 6 by 9. Also let 

g 

I 

d^e dd 

This is a linear differential equation of the second order with 
constant coefficients and can be solved by the method of §185. 
The auxiliary equation is 

m^ + 2km + w^ = 0, 
whence 

m = - k ± \/k'^ - o) 2. 

When the velocity is not very great, as in the case of an ordinary 
pendulum, k is very small for air and is much less than oj. The 
expression under the radical sign is negative. We write then 

m = - k ± i Vw^ - A;2, 
oj^ — k^ being positive. 
The solution of (3) is 

e = Ae-*' cos [t^/(a^ — k'^ - e], 
or, multiplying both sides by I and replacing Al by B, 
s = 5e~*' cos[t\/(t)^ — k- — c]. 

The motion is a damped harmonic motion. The amplitude de- 

2t 

creases with the time. The period . — is a little greater 

■yct}^ — k'^ 

2x . 

than — , the period of the free motion. 

Since k is very small in comparison with w, we can, for an 
approximate solution of our problem, neglect k^ in comparison 
with co^. Equation (4) becomes 

s = 5e-*' cos (aj< — e). (5) 

This represents the motion with a high degree of approximation. 



§186] DIFFERENTIAL EQUATIONS 353 

The arbitrary constants B and e can be determined by suitable 
initial conditions. For example, let it be given that s = sq and 

ds 

'77 = when t = 0. On differentiating (5) we obtain 

ds 

jT= Be-'"[ — k cos (cjt — e) — CO sin(co< — e)]. (6) 

For f = we obtain from (5) and (6) 

So = B cos € 
= B( — k cos e + w sin e). 

From the latter of these two equations 

tan € = — . 

CO 

From the former 



1 -\ — i = 



to the degree of approximation used above. We have then as the 
approximate equation of motion 

s = SqC"*' cos (_ut — e) (7) 

where 

k k 
€ = tan~' — = — J approximately. 

Since k is very small, c is very small. 

It follows from (5) and (7) that the period of the pendulum 
in the case just considered is very little different from that of 
the same pendulum swinging in a vacuum. The amplitude of 
the swing, however, is affected and diminishes continually with 
the time. 



23 



INDEX 

(Numbers refer to pages) 



Acceleration, 52, 83, 136, 138 

angular, 138 

average, 52 
Algebraic function, 4, 55 

definition of, 56 

rational, 6 
Alternating series, 316 
Angle, 

between two lines, 321 

between two planes, 236 
Angular acceleration, 138 

velocity, 138 
Anti-derivative, 46, 49 
Applications, 218 
Arc, 

differential of length of, 100, 
181 

length of, 218 
Area, 

by double integration, 253, 
258 

moment of inertia of, 281 

polar coordinates, 183, 218 

of surface of revolution of, 
112, 219, 274 

under a curve, 75, 218 
Arithmetic mean, 116 
Axes, coordinate, 228 

Base, 

change of, 147 
naperian, 146 
natural, 146 



147 

199 



func- 



Cable, parabolic, 82 
Catenary, 161 



Center, 

of curvature, 290, 294 

of gravity, 250, 254, 274 

of mass, 263 
Centroid, 263, 265, 274, 281 

of a line, 268 

of an area, 268 
• of a solid, 268 
Change of base, logarithm, 

of limits of integration, 
Circle, 

curvature of, 290 

of curvature, 290 
Circular and exponential 

tions, relation between. 
321 
Circular functions, 122, 321 
Comparison test, 316 
Complex numbers, 323 
Compound interest law, 155 
Computation 

of IT, 321 

of logarithms, 320 
Concavity of curve, 90 
Conicoid, 242 
Constant, 1 

Continuous function, 31 
Contraction of curve, 7 
Convergence of series, 315 
Convergent series, 303 
Co6rdinate axes, 228 

planes, 228 
Cosines, direction, 230 
Curvature, 289 

approximate formula for, 293 

center of, 290, 294 



356 



356 



INDEX 



Curvature, circle of, 290 

of a circle, 290 

defined, 289 

parametric equations, 292 

radius of, 290 
Curve, 

contaction of, 7 

direction of, 178 

elongation of, 7 

orthographic projection of, 
7 

shear of, 7 

translation of, 7 
Curves 

of hyperbolic type, 2 

maxima and minima points 
of, 20 

of parabolic type, 2 
Cylindrical surfaces, 244 

Damped harmonic motion, 351 
Definite double integrals, 255 

integrals, 88, 103, 104 
DeMoivre's Theorem, 323 
Dependent variable, 1 
Derivative, 21 

first, 21 

of a constant, 43 

of a function of a function, 66 

of a quotient, 58 

of circular functions, 122, 
124, 131 

of exponential functions, 145 

of logarithmic functions, 145 

of sin u, 122 

of the product of two func- 
tions, 57 

of the sum of a function and 
a constant, 39 

of the sum of a constant and 
a variable, 56 

of the sum of two functions, 43 



Derivative of m„, 42, 59 

second, 70 

total, 328, 329 
Derivatives of higher order, par- 
tial, 247 

partial, 246 
Differential, 88, 95 

exact, 328, 332 

of length of arc: polar 
coordinates, 181, 218 
rectangular coordinates, 
100, 218 
Differential equation, 338 

exact, 335 

linear, 342 

of higher order, 347 

order of, 338 

variables separable, 339 
Diiferentiation, 31, 97 

implicit, 44 

logarithmic, 153 
Direction cosines, 230, 239 

of curve, 178 
Distance 

between two points, 230 

of a point from a plane, 237 
Divergent series, 303 
Double integration, 251, 253, 258, 

255 
Duhamel's theorem, 105 

Element of integration, 113 

Ellipsoid, 242 

Elliptic paraboloid, 244 

Elongation of curve, 7 

Envelope, 289, 296, 336 
of normals, 300 

Equation, 

differential, 338 
exact differential, 335 
homogeneous differential, 
340 



INDEX 



357 



Equation, linear differential, 342, 
347 
of first degree in x, y and z, 

233 
of a plane, 233 

intercept form of, 234 
normal form of, 232 
Equations of a line, 238 

parametric, 67 
Evolute, 289, 294, 300 

the envelope of the normals, 
300 
Exact differential, 328, 332 

equation, 335 
Exponential functions, 9, 145, 321 
Extended law of the mean, 307 

Falling body, 22 
Family of curves, 297 
Formulas, integration, 185, 201, 

203, 205, 206, 207 
Formulas, Wallis', 204 
Fractions, partial, 211 
Function, 1 

a", 9 

a cos X + '^ sin x, 9, 165 

a cos X, 9 

ax' +px + y, 165 

6 sin X, 9 

sin X, 9 

maximum and minimum val- 
ues of, 20 

mx ± y/a" - x% 168 

xS 16 
Functions, 

algebraic, 4, 5, 56 

circular, 122, 321 

continuous and discontinu- 
ous, 31 

exponential, 145, 321 

hyperbolic, 159 

implicit, 44 



Functions, logarithmic, 145 
power, 1 
rational, 6 

integral, 6 
transcendental, 6 
transformations of, 10 

General solution of a differential 
equation, 338 

Harmonic motion, 140 

damped, 351 
Homogeneous differential equa- 
tion, 340 
Hyperbolic functions, 159 

paraboloid, 244 
Hyperboloid, 

of one sheet, 244 
of two sheets, 244 

Implicit differentiation, 44 
Improper integrals, 218, 223, 226 
Increments, 13 
Indefinite integrals, 104 
Independent variable, 1 
Indeterminate forms, 30, 31, 303, 

304 
Infinite limits of integration, 226 
Infinite series, 303 
Infinitesimals, 29, 88 

limits of ratio of two, 93 

order of, 91 
Infinity, 29 

Inflection, point of, 70 
Integral, the 

I sec' X dx, 204 
I e°* sin nx dx, 202 
I e "' cos nx dx, 202 

Integrals, improper, 218 
Integration, 46, 49, 185 



358 



INDEX 



Integration, by parts, 201 
double, 251, 253, 258 
formulas, 185, 201, 203, 

205, 206, 207 
of expressions containing, 

g-T " + bx + c,.190 

Va^ + x^ V a^ - x2, 
Vx^ - a^, 196 
powers of x and of a + hx, 

192 



. Ca smx -\- I 
of I — : 

J c sin a; + .i 



bcos x 



dx, 209 

J cos X 

of powers of trigonometric 

functions, 193 
successive, 250 
triple, 250, 260 
Intercept form of the equation of 

a plane, 234 
Inverse functions, 67 
Involutes, 302 

Law of the mean, 306 

extended, 307 
Length of arc, 

polar coordinates, 181, 218 

rectangular coordinates, 100, 
111, 218 
Limit, 

definition of, 27 

of the quotient of two 
infinitesimals, 93 

of S/(x)x, 102 
Limits, 

infinite, of integration, 226 

of integration, change of, 199 

theorems on, 30 
Line, 

direction cosines of, 230 

equations of, 238 
Linear differential equation, 

of first order, 342 

of higher order, 347 



Loci, theorems on, 10 
Logarithmic differentiation, 153 
Logarithmic functions, 145 
Logarithms, computation of, 320 

Maclaurin's series, 311 
theorem, 303, 310 

Maxima and minima, 20, 60, 165 
applications of, 174 
by limits of curve, 169 
determined by derivative, 

169 
second derivative test for, 
172 

Maximum and minimum values 
of functions, 60 

Maximum defined, 20 

Mean, arithmetic, 116 

Mean value of a function, 117 

Mean, law of the, 306, 307 

Minimum defined, 20 

Moment, 263 

Moment arm, 263 

Moment of inertia, 250, 277 
of area, 280, 281, 283 
of a solid, 284 
polar coordinates, 283 
translation of axes, 280 
with respect to a plane, 285 

Naperian base, 146 

Natural base, 146 

Normal form of the equation of a 

plane, 232 
Normal, length of, 68 
Normals, envelope of, 300 

Octant, 229 

Order of differential equation, 

338 
Orthographic projection of curve, 

7 



INDEX 



359 



Pappus, theorems of, 274 
Parabolic cable, 82 
Paraboloid, 

elliptic, 244 

hyperbolic, 244 

of revolution, 24 
Parallel planes, 236 
Parameter, 297 
Parametric equations, 67, 292 
Partial derivatives, 246 

of higher order, 247 
Partial fractions, 211 
Particular integral of a differen- 
tial equation, 338, 339 
Path of a projectile, 85 
Pendulum, the simple, 141 
Per cent, rate, 159 
Perpendicular planes, 236 
Plane, 

general equation of, 234 

intercept form of equation 
of, 234 

normal form of equation of, 
232 
Planes, 

angle between two, 236 

coordinate, 228 

parallel, 236 

perpendicular, 236 
Point of inflection, 70 
Points, distance between, 230 
Polar coordinates, 178 

area, 183, 258 

centroid, 276 

differential of arc, 181 

direction of curve in, 178 

moment of inertia in, 283 
Polynomials, 4 
Power function, 1, 33 

derivative of, 33, 35, 41, 42 

hyperbolic type, 2 

law of, 3 



Power function, parabolic type, 2 
Power series, .'505 
Projectile, path of, 85 

Quadric surface, 242 
Quotient, derivative of, 58 

Radius of curvature, 290 

approximate formula for, 293 
Radius of gyration, 278 
Rate of change, 37 
Rational algebraic function, 6 
Relative rate, 159 
Rolle's theorem, 305 

Second derivative, 70, 172 
Series, 303 

alternating, 316 

convergence of, 315 

convergent, 316, 303 

divergent, 303, 315 

infinite, 303 

Maclaurin's, 311 

power, 305 

Taylor's, 311 

test for convergence, 315 
Shear of curve, 8 
Simple harmonic motion, 140 

pendulum, 141 
Slope of tangent line, 17 
Solid geometry, 228 
Solid of revolution, 109, 218 

surface of, 219 

volume of, 109, 218, 260 
Solution of differential equation, 

338 
Subnormal, length of, 68 
Subtangent, length of, 68 
Successive integration, 250 
Surface of revolution, 112, 219 
Surfaces, 

cylindrical, 244 



360 



INDEX 



Surfaces of revolution, 240, 
274 
quadric, 242 
Symmetric form of the equations 
of a line, 238 

Tangent, 

length of, 68 

slope of, 17 
Taylor's series, 311 

theorem, 303, 308 
Tests for convergence, 316 
Theorems of Pappus, 274 
Total derivative, 328 

differential, 329 
Transcendental functions, 6 



Transformation of functions, 10 
Translation of curves, 6 
Triple integration, 250, 260 

Variable, 1 

Velocity, 16, 52, 136, 138 

average, 16 

of a falling body, 22 
Volume by triple integration, 260 

of a solid of revolution, 109, 
218, 274 

Wallis' formula, 204 
Water pressure, 114, 219 
Work done by a variable force, 
77, 107, 219 



Date Due 


QTR 

inAM oh: 


9 77 






RBTD „, 


\h 77 






G|U 
































































































































(S 


PRINTED 


IN U. S. A. 





The RALPH D. REED UBRART 



OEPAKTM KNT OF GMLOGT 

UflVERSITY of CALIFORNIA 



UC SOUTHERN REGIONAL UBRARYFAOLJTjr 

mMiiiii 

A 000 584 805 6